01_Stoichiometry & Redox Reaction_final

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Chapter-1 / Stoichiometry & Redox Reaction

Chapter 1

Stoichiometry & Redox Reaction Amedeo Avogadro

Born

09 August 1776

Died

09 July 1856

Nationality

Italy

Field

Chemistry

Known for

Avogadro’s number

Lorenzo Romano Amedeo Carlo Avogadro, (1776 - 1856) was an Italian savant chemist, most noted for his contributions to the theory of molarity and molecular weight. As a tribute to him, the number of atoms of one mole of a substance, 6.02 × 1023 is known as Avogadro's number. In honour of Avogadro's contributions to the theory of molarity and molecular weights, the number of molecules in one mole was renamed Avogadro's number, NA. It is approximately 6.0221415 × 1023. Loschmidt first calculated the value of Avogadro's number, now called Avogadro's constant, which is still sometimes referred to as the Loschmidt number in German-language countries (Loschmidt constant now has another meaning). Avogadro's number is commonly used to compute the results of chemical reactions. It allows chemists to determine the exact amounts of substances produced in a given reaction. Clausius, by his kinetic theory on gases, was able to give another confirmation of Avogadro's law. Not long after, in his researches regarding dilute solutions (and the consequent discovery of analogies between the behaviour of solutions and gases), J. H. van 't Hoff added his final consensus for the triumph of the Italian scientist, who since then has been considered the founder of the atomic-molecular theory.

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SECTION-I (For Students of Class XI)

1.1

The Concept of Atoms And Molecules The Atoms The smallest particle of an element that takes part in chemical reaction.

Dalton’s Atomic Theory The main postulates of this atomic theory are Matter is discrete (i.e., discontinuous) and is made up of atoms. An atom is the smallest

(i)

(chemically) indivisible particle of an element, which can take part in a chemical reaction. Atoms of different elements have different properties and different weights.

(ii)

(iii) Atoms cannot be created or destroyed. So a chemical reaction is nothing but a rearrangement of atoms and the same number of atoms must be present before and after the reaction. (iv)

A compound is formed by the union of atoms of one element with atoms of another in a fixed ratio of small whole numbers (1 : 1, 1 : 2, 2 : 3 etc). All the postulates of Dalton’s atomic theory have been proved to be incorrect. An atom is divisible in the sense that it has a structure. Sub-atomic particle are known as electron, proton and neutron. The existence of isotopes for most elements show that atoms of the same element need not have the same mass. The atomic weight of an element is, in fact, a mean of the atomic masses of the different isotopes of the element.

Atomic Weight An atom is so minute that it cannot be detected even with the most powerful microscope, let alone placed on a balance pan and weighed. So there is no question of determining the absolute weight of an atom. So chemists decided to determine the relative masses of atoms (i.e., how many times one atom of an element is heavier than another). Hydrogen atom was first selected as standard. i.e. Atomic weight of an element =

Weight of 1 atom of the element Weight of 1 atom of hydrogen

When we state that the atomic weight of chlorine is 35.5, we mean that an atom of chlorine is 35.5 times heavier than an atom of hydrogen. It was later felt that the standard for reference for atomic weight may be oxygen, the advantage being that the atomic weights of most other elements became close to whole numbers.

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Atomic weight of an element =

Chapter-1 / Stoichiometry & Redox Reaction

Weight of 1 atom of the element 1 × weight of 1 atom of oxygen 16

The modern reference standard for atomic weights is carbon isotope of mass number 12. Atomic weight of an element =

Weight of 1 atom of the element 1 × weigth of 1 atom of carbon − 12 12

On this basis, atomic weight of oxygen 16 was changed to 15.9994. Nowadays atomic weight is called relative atomic mass and denoted by amu (atomic mass unit). The standard for atomic mass is C12. (i) Atomic weight is not a weight but a number. (ii) Atomic weight is not absolute but relative to the weight of the standard reference element (C12). (iii) Gram atomic weight is atomic weight expressed in grams, but it has a special significance with reference to a mole. Dulong and Petit measured the specific heats of a number of metals and found that the product of the specific heat and the atomic weight is a constant, having an approximate value of 6.4. Specific heat (cal/g-deg) × atomic weight (g/g–atom) ≃ 6.4 (cal/deg.g.atom). This correlation has been used to ‘correct’ the atomic weights of some elements in the periodic table. Dulong and Petit’s law is applicable only to metals.

The Molecule Avogadro (1811) suggested that the fundamental chemical unit is not an atom but a molecule, which may be a cluster of atoms held together in some manner causing them to exist as a unit. The term molecule means the smallest particle of an element or a compound that can exist free and retain all its properties. Consider a molecule of sulphur dioxide. It has been established that it contains one atom of sulphur and two atoms of oxygen. This molecule can split up into atoms of sulphur and oxygen. So the smallest particle of sulphur dioxide that can exist free and retain all its properties is the molecule of sulphur dioxide. The term molecule is also applied to describe the smallest particle of an element which can exist free. Thus a hydrogen molecule is proved to contain two atoms; when it is split up into atoms, there will be observed a change in properties. The number of atoms of an element in a molecule of the element is called its atomicity.

Molecular Mass Molecular mass of a substance is an additive property and can be calculated by adding the atomic masses of all the atoms of different elements present in one molecule. Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail: [email protected]

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Methods of determining Molecular Masses (a)

Vapour density method V. D. =

1 Mass of volatile substance 2 Volume of vapour of its chloride at STP

× 22400

Molecular mass = 2 × V.D. (b)

Diffusion method According to Graham’s law of diffusion, rate of diffusion of a gas is inversely proportional to the square root of its molecular mass. r1 = r2

(c)

M2 M1

Victor Meyer method This method applies to volatile organic liquids. Suppose vapour of an organic liquid having mass W g occupies a volume of V mL at STP. Then its molecular mass is, Mol. mass =

W × 22,400. V

Avogadro Number (NA). The number of carbon atoms present in one gram-atom (1 mole atom) of C–12 isotope is called Avogadro’s number. One gram-atom (12 grams) of C–12 contains 6.02 × 1023 atoms. Thus the numerical value of Avogadro’s number (NA) is 6.02 × 1023 per mol. It should be noted that 1 a.m.u.=

1 th of mass of a C12 atom 12

Avogadro Hypothesis It states that equal volumes of gases at the same temperature and pressure contain equal number of molecules. It means that 1 ml of hydrogen, oxygen, ammonia, or a mixture of gases taken at the same temperature and pressure contains the same number of molecules. Application (a)

To prove that simple elementary gas molecules are diatomic Consider the experimental result, 1 volume of hydrogen + 1 volume of chlorine  → 2 volumes of hydrogen chloride at the same temperature and pressure 1 volume contains ‘n’ molecules. Then n molecules of hydrogen + n molecules of chlorine  → 2n molecules of hydrogen chloride. Cancelling the common ‘n’, we have 1 molecule of hydrogen + 1 molecule of chlorine  → 2 molecules of hydrogen chloride.

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Chapter-1 / Stoichiometry & Redox Reaction

A molecule of hydrogen chloride should contain at least 1 atom of hydrogen and 1 atom of chlorine. Two molecules of hydrogen chloride should contain at least 2 atoms of hydrogen and 2 atoms of chlorine and these should have come from 1 molecule of hydrogen and 1 molecule of chlorine respectively. Thus Avogadro’s hypothesis enables us to establish that hydrogen and chlorine molecules must contain at least 2 atoms (Measurement of the ratio of specific heats of these gases at constant pressure and at constant volume, CP/CV = 1.4, establishes that these gases are diatomic). (b)

[To establish the relationship between molecular weight and vapour density of a gas. The absolute density of gas is the weight of 1 litre (dm3) of the gas at S.T.P. [Standard Temperature (0°C and Pressure (1 atmosphere)]. The relative density or vapour density of gas = Vapour density of a gas = =

Density of the gas Density of hydrogen

Weight of 1 litre of gas at STP Weight of 1 litre of hydrogen at STP

Weight of a certain volume of the gas Weight of the same volume of hydrogen at the same temperature and pressure

So the vapour density of a gas is defined as the ratio of the weight of a certain volume of the gas to the weight of the same volume of hydrogen at the same temperature and pressure. ∴

Vapour density (V.D.) of a gas =

Weight of 'V ' litres of the gas Weight of 'V ' litres of hydrogen at the same temperature and pressure

Let ‘V’ litres of the gas contains ‘n’ molecules.

∴

V.D. of a gas =

Weight of 'n ' molecules of the gas Weight of 1 molecule of the gas = Weight of 'n ' molecules of hydrogen Weight of 1 molecule of hydrogen

V.D. of a gas =

Weight of 1 molecule of the gas 1 Weight of 1 molecule of the gas = × Weight of 2 atoms of hydrogen 2 Weight of 1 atom of hydrogen

V.D. of a gas =

1 × Molecular weight of the gas 2

Molecular weight of the gas = 2 × Vapour density of the gas.

The volume occupied by a gram molecular weight of any gas is called a molar volume and

(c)

it is 22.4 L at STP.

1.2

Mole Concept

A mole of any substance is related to : (a)

mass of a substance

(b)

number of particles

(c)

volume of the gaseous substance Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail: [email protected]

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Mole-Particle Relationship A mole is a collection of 6.023 × 1023 particles, ions, atoms etc. i.e.

(i)

6.023 × 1023 atoms of Na constitute one mole atom of Na.

(ii)

6.023 × 1023 molecules of oxygen constitutes 1 mole of oxygen molecules.

(iii) 6.023 × 1023 electrons constitute one mole of electrons.

Mole-Weight Relationship One mole of every substance weighs equal to the gram atomic weight of the substance or to the gram molecular weight of the substance. e.g. (i) 1 mole of sodium weighs 23 g of Na. (ii) 1 mole of CaCO3 weighs 100 g.

Mole-Volume Relationship One mole of every gas occupies 22.4 lit. of volume at STP. i.e.

1 mole of O2 occupies 22400 ml of volume at STP. 1 mole of He occupies 22400 ml of volume at STP.

Illustrations Illustration 1 Determine the number of moles of H2 in 0.224 litre of hydrogen gas at STP (assuming ideal gas behaviour). Solution 22.4 litres of H2 at STP contain 1 mole of H2. 0.224 litre of H2 at STP has = 0.01 mole Illustration 2 What will be the number of moles in 13.5 g of SO2Cl2 ? Solution Molecular mass of SO2Cl2 = 32 + 32 + 71 = 135 g mol–1 No. of moles =

Mass in gram 13.5 = = 0.1 mole gram mol. wt. 135

Illustration 3 What will be the number of moles of oxygen in one litre of air containing 21% oxygen by volume under standard conditions. Solution 100 ml of air at STP contains 21 ml of O2

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1000 ml of air at STP contains = ∴

No. of moles = =

Chapter-1 / Stoichiometry & Redox Reaction

21 × 1000 = 210 ml O2 100

Vol. in litres under STP conditions 22.4 litre 210 = 0.009 mole. 22400

Practice Exercise 1.

Calculate the total number of electrons present in 1.6 g of methane. Molecular wt. of methane = 16 a.m.u.

2.

0.24 gm of a volatile substance upon vaporisation gives 45 ml vapours at STP. What will be the vapour density of the substance ?

Answers 1. 6.023 × 1023

1.3

2. 59.

Concept of Limiting Reagent

In reactions involving more than one reactant, one has to identify first, of all the reactant, which is completely consumed (limiting reagent), one can identify the limiting reagent as follow N 2 + 3H 2  → 2NH 3 Initial mole

5

12

−

If N2 is the limiting reactant moles of NH3 produced = 10. If H2 is the limiting reactant moles of NH3 produced =

3 × 12 = 8 . The reactant producing the least number of mole of the product is 2

the limiting reactant, hence H2 is the limiting reactant. The limiting reactant can also be ascertained by knowing the initial number of equivalents or milli equivalents of each reactant. The reactant with least number of equivalents or milli equivalents is the limiting reactant. The equivalent methods to identify the limiting reactant used not require balancing of chemical equation.

1.4

Gravimetric Analysis Calculation involving in mass – mass relationship In general, the following steps are adopted in making necessary calculations

1.

Write down balanced molecular equation for the chemical change

2.

Write down the no of moles below the formula of each of the reactant and product

3.

Write down the relative masses of the reactants and the products with the help of formula below the respective formula. These shall be the theoretical amounts of reactants and product.

4.

By the applications of unitary method, mole concept or proportionality method, the unknown factor or factors are determined. Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail: [email protected]

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Calculation involving percent yield In general, when a reaction is carried out in the laboratory we do not obtain actually the theoretical amount of the product. The amount of the product that is actually obtained is called the actual yield. Knowing the actual yield and theoretical yield the percentage yield can be calculated as % yield =

Actual yield × 100 Theoretical yield

Weight-Volume Relationship 2Mg(s) + O 2 (g)  → 2MgO(s)

In the above reaction, one can find out the volume of O2 at STP required to react with 10 gm of 10 . The moles of O2 required would be 1/2 the moles of Mg. Therefore 24 1 10 1 10 moles of O 2 = × . Since 1 mole of gas (ideal) occupies 22.4 L at STP, therefore × moles of 2 24 2 24 1 10 × 22.4 L = 4.67 L. O2 would occupy, × 2 24

Mg. The moles of Mg is

Volume-volume Relationship Let us consider the reaction H2(g) + ½ O2(g) → H2O(l). We are given 10 L of H2 at a given temperature and pressure. How many litres of O2 would react with hydrogen at the same temperature and pressure? From the ideal gas equation [PV = nRT] it is clear that the volume of an ideal gas is directly proportional to its no. of moles. Therefore under the same conditions of P and T, VH2 VO2

=

n H2 n O2

. Since the molar ratio is 2 : 1 (H2 : O2), ∴ the volume ratio would also be 2 : 1.

Therefore the volume of O2 required would be 5L. On the other hand if we need to calculate the volume of O2 at a different T and P, then PH2 VH2 = n H 2 R TH2 , PO2 VO2 = n O2 RTO2 , and dividing we get ∴

VO2 =

PH2 VH2 PO2 VO2

PH 2 VH2 PO2

×

=

n H 2 TH2 n O2 TO2

n O2 TO2 n H2 TH2

,

.

1.5 Concentration Units There are several methods to express the concentration of solution. We are going to describe some very important terms used to represent the concentration of solution. (a)

Percent by weight

(b)

Percent by volume

(c)

Percent by weight/volume -8-

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(d)

Mole fraction

(e)

Molarity

(f)

Molality

(g)

Normality

(h)

ppm

Chapter-1 / Stoichiometry & Redox Reaction

Percent by Weight (w/w) It is the weight of the solute as a per cent of the total weight of the solution. That is, weight of solute %by weight of solute = × 100 weight of solution For example, if a solution of HCI contains 36 % HCI by weight, it has 36 g of HCI in 100 g of solution.

Percent by Volume (v/v) % by volume =

volume of solute × 100 volume of solution

For example, it we have 35% C2H5OH solution by volume means 25 ml C2H5OH is present per 100 ml of the solution.

Percent by weight/volume (w/v) %=

weight of solution × 100 volume of solution

i.e.30% HCl solution, means 30 g of HCl is present per 100 ml of solution

Mole Fraction (X) A simple solution is made of two substances; one is the solute and the other is solvent. Mole fraction, X, of solute is defined as the ratio of the number of moles of solute and the total number of moles of solute and solvent. Thus,

X solute =

moles of solute ( moles of solute + moles of solvent )

If n represents moles of solute and N number of moles of solvent, n X solute = n+N Notice that mole fraction of solvent would be N X solvent = n+N Mole fraction is unitless and ( X solute + X solvent ) = 1

Molarity (M) In current practice, concentration is most often expressed as molarity. Molarity is defined as the number of moles of solute per litre of solution. Molarity =

moles of solute w/M w × 1000 or M = = v / 1000 M′V volume in litres

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w = Mass of solute in grams M = molecular weight of solute in gm/mol. V = value of solution in ml. It’s unit is mol/L.

Molality (m) Molality of a solution is defined as the number of moles of solute per kilogram of solvent: moles of solute Molality = mass of solvent in ki log rams w w × 1000 M M= = W MW 1000

or where

w = mass of solute in grams M = molecular wt of solute W = mass of solvent in gram.

Normality (N) Normality of a solution is defined as number of equivalents of solute per litre of the solution: w Number of gram equivalents of solute w × 1000 Normality = or N = E = V VE Volume of solution in litres 1000 w = mass of solute in gram V = volume of solution in ml E = equivalent wt of solute

ppm (parts per million) It is the mass of solute in grams present per 106 grams of solution.  weight of solute  = × 10 6  ppm  weight of solution   volume of particular gas  and also  × 10 6  ppm  volume of gaseous mixture 

Illustrations Illustration 1 A solution of NaCl 0.5% by wt. If the density of the solution is 0.997 g/ml, calculate the molality, molarity, normality and mole fraction of the solute.

Solution Number of moles of NaCI = -10-

mass of NaCI Molecular mass of NaCI

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=

Chapter-1 / Stoichiometry & Redox Reaction

0.5 = 0.0085 58.5

By definition, Molality =

No. of moles of solute × 1000 0.00854 × 1000 = = 0.086m 99.5 Mass of solvent in grams

Volume of the solution =

Mass of solution in grams Density in gm per ml

100 = 100.3 0.997 Nimber of moles of soloute Now Molarity = Volume of solution in litres =

0.00854 × 1000 = 0.08514M 100.3 Number of gram equivalents of solute and Normality = volume of solution in litres

=

=

0.5 × 1000 100.3 × 58.5

= 0.0852 To calculate mole fraction of the solute 99.5 No. of moles of water in 99.5g = = 5.5277 18 0.5 Moles of NaCl = = 8.547 × 10 −3 58.5 n NaCl X NaCl = n NaCl + n H2O 8.547 × 10 −3 = 1.5433 × 10 −3 M 5.5277 + 8.547 × 10 −3 = 1 − X NaCl = 0.9984

=

X H2 O

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SECTION-II (For Students of Class XII)

1.6

Law of Equivalents

Before heading for the law of equivalents, let us first discuss certain definitions.

Molarity (M) It is defined as the number of moles of solute present in one litre of solution. Molarity (M) =

number of moles of solute Volume of solution (in litres)

Let the weight of solute be w g, molar mass of solute be M1g./mol and the volume of solution be V litre. Number of moles of solute =

weight of solute w = Atomic or molar mass of solute M1

w 1 × M1 V(in litres)

∴

M=

∴

Number of moles of solute =

w = M × V (in litres) M1

Normality (N) It is defined as the number of moles of equivalents of a solute present in one litre of solution. Equivalent is also the term used for amount of substance like mole with the different that one equivalent of a substance in different reactions may be different as well as the one equivalent of each substance is also different. Normality (N) =

number of equivalents of solute Volume of solution (in litres)

Let the weight of solute be w g, equivalent mass of solute be E g/eqv. And the volume of solution be V litre. Number of equivalents of solute =

weight of solute w = Equivalent mass of solute E

w 1 × E V(in litres)

∴

N=

∴

Number of equivalents of solute = N × V (in litre)

Equivalent Mass Equivalent mass = ∴

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Atomic or molecular mass M1 = 'n ' factor n

Number of equivalents of solute =

w w w×n = = E M1 / n M1

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∴

Chapter-1 / Stoichiometry & Redox Reaction

Number of equivalents of solute = n × number of moles of solute

Also, N =

w 1 w 1 × = × ×n M1 / n v (in litre) M1 V (in litre)

N=M×n ∴

Normality of solution = n × molarity of solution.

Illustrations Illustration 4 1.60 g of a metal were dissolved in HNO3 to prepare its nitrate. The nitrate was strongly heated to give 2 g oxide. Calculate equivalent weight of metal. Solution ∆ 3 M  → M(NO 3 )n  → M 2 On HNO

where n is valency of metal ∴ neq. of metal = neq. of nitrate = neq. of metal oxide = neq. of oxygen w metal E metal

=

1.60 E

=

w oxygen Eoxygen 2 − 1.60 8

E = 32 Illustration 5 1.0 g of metal nitrate gave 0.86 g of metal sulphate. Calculate the equivalent weight of metal. M(NO3)n → M2(SO4)n ∴

neq. of M(NO3)n = neq. of M2(SO4)n 1 0.86 = EM + 62 E + 48 1 0.86 = E + 62 E + 48

∴

E = 38

Illustration 6 1.5276 g of CdCl2 was found to contain 0.9367 g of Cd. Calculate atomic weight of Cd. Weight of Cd = 0.9367 g ∴

Weight of Cl = 1.5276 – 0.9367 = 0.5909 g For CdCl2; neq. of cadmium = neq. of chlorine 0.9367 0.5909 = E 35.5

∴ ∴

E = 56.275 Atomic weight = Eq. wt. × Valency = 56.275 × 2 = 112.55.

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Practice Exercise 3.

1.8 g of an element displaces 2.04 g copper from CuSO4 solution. If equivalent weight of Cu = 31.7, what is equivalent weight of iron ?

4.

2 g of a metal in H2SO4 gives 4.51 g of the metal sulphate. The specific heat of metal is 0.057 cal/g. Calculate the valency and atomic weight of metal.

5.

A hydrated sulphate of metal contained 8.1% metal and 43.2% SO4–2 by weight. The specific heat of metal is 0.24 cal/g. What is hydrated sulphate ?

Answers 3. 27.97

4. 114.72

5. M2(SO4)3.18H2O

Dilution Effect When a solution is diluted, the moles and equivalents of solute donot change but molarity and normality changes while on taking out a small volume of solution from a larger volume, the molarity and normality of solution do not change but moles and equivalents change proportionately. In stoichiometry, the biggest problem is that for solving a problem we need to know a balanced chemical reaction. Since the number of chemical reactions are too many, it is not possible to remember all those chemical reactions. So, there is need to develop an approach which does not require the use of balanced chemical reaction. This approach makes use of a law called law of equivalence. The law of equivalence provide us the molar ratio of reactants and products without knowing the complete balanced reaction, which is as good as having a balanced chemical reaction. The molar ratio of reactants and products can be known by knowing the n-factor of relevant species. According to the law of equivalence, whenever two substances react, the equivalents of one will be equal to the equivalents of other and the equivalents of any product will also be equal to that of the reactant. Let us suppose we have a reaction, A + B → C + D. In this reaction, the number of moles of electrons lost by 1 mole of A are x and the number of mole of electrons gained by 1 mole of B are y. Since, the number of mole of electrons lost and gained are not same, the molar ratio in which A & B react cannot be 1 : 1. Thus, if we take y moles of A, then the total moles of electrons lost by y moles of A would be (x × y). Similarly, if x moles of B are taken, then the total mole of electrons gained by x moles of B would be (y × x). Thus, the number of electrons lost by A and number of electrons gained by B becomes equal. For reactant A, its n-factor is x and the number of moles used are y. So, The equivalents of A reacting = moles of A reacting × n-factor of A = y × x. Similarly, for reactant B, its n-factor is y and the number of moles used are x. So, -14-

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Chapter-1 / Stoichiometry & Redox Reaction

The equivalents of B reacting = moles of B reacting × n-factor of B =x×y Thus, the equivalents of A reacting would be equal to the equivalents of B reacting. Thus, the balancing coefficients of the reactant would be as yA (n − factor = x)

+

xB

(n − factor = y)

 → C +D

The n-factor of A & B are in the ratio of x : y, and their molar ratio is y : x. Thus, molar ratio is inverse of the n-factor ratio. In general, whenever two substances react with their n-factors in the ratio of a : b, then their molar ratio in a balanced chemical reaction would be b : a. To get the equivalents of a substance, its n-factor is to be known. Let the weight of the substance used in the reaction be w g. Then, equivalents of substance reacted would be

W w w or = × n (where E and E M1 / n M1

M1 are the equivalent mass and molar mass of the substance) Thus, in order to calculate the equivalents of a substance, knowledge of n-factor is a must.

Illustrations Illustration 7 Calculate normality and molarity of the following? (a) 0.74 g of Ca(OH)2 in 5 ml of solution. (b) 3.65 g of HCl in 200 ml of solution. (c) 1/10 moles of H2SO4 in 500 ml of solution. Solution (a)

Eq. of Ca(OH)2 =

0.74 74 / 2

w   Eq. =  E 

Volume of solution = 5/1000 litre ∴ N=

0.74 × 1000 × 2 74 × 5

N= 4 ∴ M = (b)

Eq. of HCl =

N 4 = =2 Total charge on anion or cation 2

3.65 36.5

and Volume of solution = 200 / 1000 litre ∴

N=

3.65 × 1000 = 0.5 36.5 × 200

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M=

and

N 0.5 = = 0.5 Valency 1

(c) Eq. of H2SO4 =

1 × 2 ( Meq. = N × V in ml) 10

= 0.2 Molar H2SO4 = 0.4 Normal H2SO4. Illustration 8 What volume of water is required to make 0.20 N solution from 1600 ml of 0.2050 N solution. Solution Meq. of conc. solution = 1600 × 0.2050 = 328 Let after dilution volume becomes V ml Meq. of dil. solution = 0.20 × V ∴

328 = 0.20 × V

∴

V = 1640 ml

Thus volume of water used to prepare 1640 ml of 0.20N solution = 1640 – 1600 = 40ml Illustration 9 What is the normality and nature of a mixture obtained by mixing 0.62 g of Na2CO3 .H2O to 100 ml of 0.1 N H2SO4 ? Solution neq. of

=

0.62 × 1000 = 0.01 62

neq. of H2SO4=

100 × 0.1 = 0.01 1000

Na 2 CO3 n eq. added Meq. left

∴

N Na 2SO4 =

+ H 2 SO4

0.01 0

0.01 0

→ Na 2 SO4 0 0.01

+H 2 O + CO 2 ↑ 0 0.01

0 0.01

0.01 × 1000 = 0.1 100

Solution becomes neutral since both acid and base are used up and Na2SO4 does not show hydrolysis. Illustration 10 How much AgCl will be formed by adding 200 ml of 5N HCl to a solution containing 1.7 g AgNO3 ?

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Chapter-1 / Stoichiometry & Redox Reaction

Solution AgNO3 + HCl → 1.7 200 × 5 170 =10−2 = 1 eq

n eq. initially

n eq. after reaction

0

AgCl + HNO3 0

0

0

1 − 0.01

0

.01

.01

∴

neq. of AgCl formed = 0.01

∴

Amount of AgCl formed = 0.01 × 143.5 = 1.435 g

∴

wAgCl = 1.435 g.

Illustration 11 Find the weight of H2SO4 in 1200 ml of a solution of 0.2N strength. Solution neq. of H2SO4 =

0.2 × 1200 = 0.24 1000

( Meq. = N × V in ml)

amount of H2SO4 = neq. × neq. wt. of H2SO4 = 0.24 × 49 = 11.76 g w = 11.76g

Practice Exercise 6.

The density of 3M solution of Na2S2O3 is 1.25g ml–1. Calculate (a) the % by weights of Na2 S2 O3 (b) mole fraction of Na2 S2 O3 (c) the molalities of Na+ and S 2 O3−2 ions.

7.

A sample of H2SO4 (density 1.787g ml–1) is labelled as 86% by weight. What is molarity of acid ? What volume of acid has to be used to make 1 litre of 0.2M H2SO4?

8.

What weight of AgCl will be precipitated when a solution containing 4.77 g NaCl is added to a solution of 4.77 g of AgNO3 ?

9.

How much BaCl2 would be needed to make 250 ml of a solution having same concentration of Cl– as the one containing 3.78 g of NaCl per 100 ml ?

Answers 6. (a) 37.92 (b) 0.065 (c) 3.865

7. 12.75 ml

8. 4.87 g

9. 16.80

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1.7 ‘n’ Factors In Non Redox Change (i)

n-factor for element: Valency of the element

(ii)

For Acids: Acids will be treated as species which furnish H+ ions when dissolved in a solvent. The n factor of an acid is the no. of acidic H+ ions that a molecule of the acid would give when dissolved in a solvent (Basicity). e.g. for HCl (n = 1), HNO3 (n = 1), H2SO4 (n = 2), H3PO4 (n = 3) and H3PO3 (n = 2)

(iii) For Bases: Bases will be treated as species which furnish OH– ions when dissolved in a solvent. The n factor of a base is the no. of OH– ions that a molecule of the base would give when dissolved in a solvent (Acidity). e.g. NaOH (n = 1), Ba(OH)2 (n = 2), al(OH)3 (n = 3), etc. (iv)

For Salts: A salts reacting such that no atom of the salt undergoes any change in oxidation state. e.g. 2AgNO3 + MgCl2 → Mg(NO3)2 + 2AgCl In this reaction it can be seen that the oxidation state of Ag, N, O, Mg and Cl remains the same even in the product. The n factor for such a salt is the total chare on cation or anion.

Illustrations Illustration 12 Calculate the n-factor in the following chemical changes. +

(ii)

H2O KMnO 4  → Mn 4 +

OH KMnO 4  → Mn 6+

(iv)

H K 2 Cr2 O7 → Cr 3+

C 2 O 24 −  → CO 2

(vi)

FeSO4 ¾→ Fe2O3

(i)

H KMnO 4 → Mn 2 +

(iii) (v)

−

+

(vii) Fe2O3 ¾→ FeSO4 Solution In this reaction, KMnO4 which is an oxidizing agent, itself gets reduced to Mn2+ under acidic

(i)

conditions. n = |1 × (+ 7) – 1 × (+ 2)| = 5 (ii)

In this reaction, KMnO4 gets reduced to Mn4+ under neutral or slightly (weakly basic conditions. n = |1 × (+ 7) – 1 × (+ 4)| = 3

(iii) Here KMnO4 gets reduced to Mn6+ under basic conditions. n = |1 × (+ 7) – 1 × (+ 6) = 1

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(iv)

Chapter-1 / Stoichiometry & Redox Reaction

Here K2Cr2O7 which acts as an oxidizing agent gets reduced to Cr+3 under acidic conditions. (It does not react under basic conditions). n = +1 × (+7) – 1 × (+ 6)| = 1

(v)

C2O42– (oxalate ion) gets oxidized to CO2 when it is reacted with an oxidizing agent. N = |2 × (+ 3) – 2 × ( 4) = –2

(vi)

Ferrous ions get oxidized to ferric ions n = |1 × (+ 2) –1 × (+ 3) = 1

(vii) Here ferric ions are getting reduced to ferrous ions N = |2 × (+ 3) – 2 × (+ 2)| = 2 Illustration 13 Predict the molar ratio in which the following two substances would react if they are assumed to be salts Ba3(PO4)2 and AlCl3

Solution N factor of Ba3 (PO4) = 3 × (+2) = 6 = n1 While n factor of AlCl3 = 1 × (+ 3) = 3 = n2 n1 6 = n2 3

If

n1 x = n2 y

Molar ratio = ∴

y (inverse of equivalent ratio) x

molar ratio in which Ba3 (PO4) and AlCl3 will react = 3 : 6 = 1 : 2.

Illustration 14 Find the n-factor for the reactants in each of the following cases (i)

I– → I2

(iii) S2O32– → S4O62–

(ii)

I2 → I–

(iv)

IO3– → Cu2+ + SO2

(ii)

n factor = 2 × |–1 – 0| = 2

(iv)

n factor = 1 × |1 – 5| = 4

Solution (i)

n factor = 1 × |0 – (–1)| = 1

(iii) n factor = 2 × |2.5 – 2| = 1 Illustration 15

Calculate the n-factor in the following redox change FeC2 O 4  → Fe3+ + CO 2

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Solution Here Fe2+ is getting oxidized to Fe3+ and C3+ is getting oxidized to C4+. The n factor w.r.t. Fe is +1 and w.r.t. C2O42– is 2 (as calculated earlier). Therefore total n factor is 3. In Redox Change For oxidizing agent or reducing agent n-factor is the change in oxidation number per mole of the substance. If we have a salt which react in a fashion that atoms of one of the element are getting

(i)

oxidized and the atoms of another element are getting reduced and no other element on the reactant side is getting oxidized or reduced, than the n–factor of such a salt can be calculated either by taking the total number of moles of electrons lost or total number of mole of electrons gained by one mole of the salt. For example, decomposition reaction of KClO3 is represented as +5 −2

−1

0

KClO 3  → KCl + O 2

In this reaction, O2– is getting oxidized to O2 and Cl+5 is getting reduced to Cl–1. In each case, 6 mole of electrons are exchanged whether we consider oxidation or reduction. n-factor of KClO3 considering oxidation = |3 (–2) – 3 (0)| = 6 or (ii)

n–factor of KClO3 considering reduction = |1 × (+5) –1 × (–1)| = 6.

Disproportionation reactions in which moles of compound getting oxidsed and reduced are not same i.e. moles of oxidizing agent and reducing agent are not same. For example, 6Br2 + 12 OH¯  → 10 Br¯ + 2BrO−3 + 6H2O In this reaction, the mole of electrons lost by the oxidation of some of the moles of Br3 are same as the number of mole of electrons gained by the reduction of rest of the moles of Br3. Of the 6 moles of Br2 used, one mole is getting oxidized, loosing 10 electrons (as reducing agent) and 5 moles of Br2 are getting reduced and accepts 10 moles of electrons (as oxidizing agent). Br2

2Br+5 + 10e¯

5Br2 + 10 e¯

Br2

+

10 Br¯

5Br2

10 Br¯ + 2Br+5

Thus, n-factor of Br2 acting as oxidizing agent is 2 and that Br2 acting as reducing agent has n-factor 10. Or when the reaction is written as 6Br2  → 10Br − + 2Br +5 -20-

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Chapter-1 / Stoichiometry & Redox Reaction

where, Br2 is not distinguished as how much of it functions as oxidizing agent and how much as reducing agent, then for calculating n-factor of compound in such reactions, first find the total number of mole of electrons exchanged (lost or gained) using the balanced equation and divide it with the number of mole of Br2 involved in the reaction to get the number of mole of electrons exchanged by one mole of Br2. In the overall reaction, the number of mole of electrons exchanged (lost or gained) is 10 and the moles of Br2 used in the reaction are 6. Thus, each mole of Br2 has exchanged 10/6 or 5/3 mole of electrons. Therefore, the n-factor of Br2 when the reaction is written without segregating oxidizing and reducing agent is 5/3. 6Br2

(n = 5 / 3)

 → 10Br − + 2Br +5 (n =1)

(n = 5)

1.8 Application of The Law of Equivalence Simple titration In this, we can find the concentration of a substance with the help of the concentration of another substance which can react with it. For example: Let there be a solution of a substance A of unknown concentration. We are given another substance B whose concentration is known (N1). We take a certain known volume of A in a flask (V2) and then we add B to A slowly till all the A is consumed by B. (This can be known with the help of indicators). Let us assume that the volume of B consumed is V1. According to the law of equivalents, the number of gm equivalents of A is equal to the number of gm equivalents of B. ∴

N1V1 = N2V2, where N2 is the concentraton of a.

From this we can calculate the value of N2.

Acid Base titration In this type of titration, the concentration of an acid in a solution is estimated by adding solution of standard base and vice versa. The equivalence point is detected by adding a few drops of a suitable indicator to the solution whose concentration is to be estimated. An acidbase indicator gives different colours with acids and bases. The choice of indicator in a particular titration depends on the pH-range of the indicator and the pH change near the equivalence point. For example. (i)

Strong Acid-Strong Base Titration In the titration of HCl Vs NaOH, the equivalence point lies in the pH-range of 4–10. Thus, methyl red (pH-range 4.2 to 6.3), methyl orange (pH-range 3.1 to 4.4) and phenolphthalein (pH-range 8.3 – 10) are suitable indicators of such titrations.

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(ii)

Weak Acid–Strong Base Titration In the titration of CH3COOH and NaOH, the equivalence point lies between 7.5 and 10. Thus, phenolphthalein is the suitable indicator.

(iii) Weak Base-Strong Acid Titration In the titration of NH4OH and HCl, the equivalence point lies in the pH range of 4 to 6.5. Thus, methyl orange and methyl red are suitable indicators. (iv)

Weak Acid-Weak Base Titration In the titration of CH3COOH and NH4OH, the equivalence point lies between 6.5 and 7.5 and the pH change is not sharp at the equivalence point. Thus, no simple indicator can be employed to detect the equivalence point.

Iodometric and Iodimetric titration The reduction of free iodine to iodide ions and oxidation of iodide ions to free iodine occurs in these titration’s. I 2 + 2e−  → 2I− (reduction) 2I −  → I 2 + 2e− (oxidation)

These are divided into two types Iodometric titration: In iodometric titrations, an oxidizing agent in allowed to react in

(i)

neutral medium or in acidic medium with excess of potassium iodide to liberate free iodine. KI + oxidizing agent → I2 Free iodine is titrated against a standard reducing agent usually with sodium thiosulphate Halogen, dichroamtes, cupric ion, peroxides, etc can be estimated by this method. I2 + 2Na2S2O3 → 2NaI + Na2S4O6 2CuSO4 + 4KI → Cu2I2 + 2K2SO4 + I2 K2Cr2O7 + 6KI + 7H2SO4 → Cr2(SO4)3 + 4K2SO4 + 7H2O + 3I2 (ii)

Iodimetric titration: These are the titrations in which free iodine is used as it is difficult to prepare the solution of iodine (volatile and less soluble in water) it is dissolved in KI solution. KI + I2 → KI3 (Potassium triiodide) This solution is first standardizes before use with the standard solution of I2 substance such as sulphite, thiosulphate, arsenite etc. are estimated. The iodimetric and iodometric titrations, starch solution is used as indicator. Starch solution gives blue or violet colour with iodine. At the end point the blue or violet colour disappears when iodine is completely changed to iodide.

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Back titration Back titration is used to calculate % purity of a sample. Let us assume that we are given an impure solid substance C weighing w gms and we are asked to calculate the percentage of pure C in the sample. We will assume that the impurities are inert. We are provided with two solutions A and B, where the concentration of B is known (N1) and that of A is not known. This type of titration will work only if the following conditions are satisfied A, B and C should be such compounds that A and B can react with each other, A and C should not react with B. Now we take a certain volume of A in a flask (the A taken should be such that the gm equivalents of A taken should be ≥ gm equivalents of C in the sample. (This can be done by taking A in excess). Now we perform a simple titration using B. Let us assume that the volume of B used is V1. In another beaker, we again take the solution of A in the same volume as taken earlier. Now, C is added to this and after the reaction is complete, the solution is being titrated with B. Let us assume that the volume of B used up is V2. Gram equivalents of B used in the first titration = N1V1 ∴

gm. Equivalents fo A inititally = N1V1 gm. Equivalents of B used in the second titration is N1V1

∴

gm. Equivalents of A left in excess after reacting with C = N1V2 gm. Equivalents of A that reacted with C = N1V1 – N1V2 gm. Equivalents of pure C = N1V1 – N1V2.

If the n factor of C is x, then the moles of pure C =

N1V1 − N1V2 x

∴

the weight of C =

N1V1 − N1V2 × Molecular weight of C. x

∴

percentage of C =

N1V1 − N1V2 Molecular wt. of C × × 100 . x w

Double titration The method involves two indicator (indicators are substance that change their colour when a reaction is complete) phenolphthalein and methyl orange. This is a titration of specific compounds. Let us consider a solid mixture of NaOH, Na2CO3 and inert impurities weighing w g. You are asked to find out the % composition of mixture. You are also given a reagent that can react with the sample, say, HCl along with its concentration (M1). We first dissolve the mixture in water to make a solution and then we add two indicators in it, namely phenolphthalein and methyl orange. Now, we titrate this solution with HCl. NaOH is a strong base while Na2CO3 is a weak base. So it is safe to assume that NaOH reacts with HCl first, completely and only then does Na2CO3 react. NaOH + HCl → NaCl + H2O Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail: [email protected]

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Once NaOH has reacted, it is the turn of Na2CO3 to react. It reacts with HCl in two steps. Na2CO3 + HCl → NaHCO3 + NaCl and then, NaHCO3 + HCl → NaCl + CO2 + H2O As can be seen, when we go on adding more and more of HCl the pH of the solution keeps on falling. When Na2CO3 is converted to NaHCO3, completely, the solution is weakly basic due to the presence of NaHCO3 (which is a weaker base as compared to Na2CO3). It this instant phenolphthalein changes colour since it requires this weakly basic solution to change its colour. Therefore, remember that phenolphthalein changes colour only when the weakly basic NaHCO3 is present. As we keep adding HCl, the pH again falls and when also the NaHCO3 reacts to form NaCl, CO2 and H2O the solution becomes weakly acidic due to the presence of the weak acid H2CO3(CO2 + H2O). At this instance methyl orange changes colour since it requires this weakly acidic solution to do so. Therefore, remember methyl orange changes colour only when H2CO3 is present. Now, let us assume that the volume of HCl used up for the first and the second reaction. i.e., NaOH + HCl → NaCl + H2O and Na2CO3 + HCl → NaHCO3 + NaCl be V1 (this is the volume of HCl from the beginning of the titration up to the point when phenolphthalein changes colour). Let the volume of HCl required for the last reaction, i.e., NaHCO3 + HCl → NaCl + CO2 + H2O be V2 (this is the volume of HCl from the point where phenolphthalein had changed colour upto the point when methyl orange changes (colour). Then, moles of HCl used for reacting with NaHCO3 = moles of NaHCO3 reacted = M1V2 moles of NaHCO3 produced by the Na2CO3 = M1V2 moles of Na2CO3 that gave M1V2 moles of NaHCO3 = M1V2 Mass of Na2CO3 = M1V2 × 106 % Na2CO3 =

M1V2 × 106 × 100 w

moles of HCl used for the first two reactions = M1V1 moles of Na2CO3 = M1V2 moles of HCl used for reacting with Na2CO3 = M1V2 moles of HCl used for reacting with only NaOH = M1V1 – M1V2 ∴

moles of NaOH = M1V1 – M1V2 Mass of NaOH = (M1V1 – M1V2) × 40

% NaOH =

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(M1V1 − M1V2 ) × 40 × 100 w

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Illustrations Illustration 16 50 mL of a solution, containing 1 g each of Na2CO3, NaHCO3 and NaOH, was titrated with 1(N) HCl. What will be titre reading if (a) only phenolphthalein is used as indicator (b) only methyl orange is used as indicator from the very beginning? (c) methyl orange is added after the first end point with phenolphthalein? Solution (a) The titration reactions in this case are Na2CO3 + HCl → NaHCO3 + NaCl and

NaOH + HCl → NaCl + H2O

Thus, we have m.e. of Na2CO3 + m.e. of NaOH = m.e. of v1 mL (say) of 1(N) HCl 1 1 × 1000 + × 1000 = 1 × v1 ; 106 40

∴ v1 = 34.4 mL.

(b) The reactions in this case are, Na2CO3 + HCl → NaHCO3 + NaCl NaHCO 3 + HCl → NaCl + H 2 O + CO 2 (produced)

NaHCO 3 + HCl → NaCl + H 2 O + CO 2 (originally present)

and NaOH + HCl → NaCl + H2O Thus, we have, m.e. of Na 2 CO 3 + m.e. of NaHCO3 + m.e. of NaHCO 3 + m.e. of NaOH (produced)

(originally present)

= m.e. of v2 mL (say) of N HCl 1 1 1 1 × 1000 + × 1000 + × 1000 + × 1000 = 1 × v 2 106 106 84 40

∴

v2 = 55.8 mL.

(c) The reaction in this case are NaHCO3 (produced) + HCl → NaCl + H2O + CO2 and NaHCO3 (originally present) + HCl → NaCl + H2O + CO2 Thus we have, m.e. of NaH 2 CO3 + m.e. of NaHCO 3 = m.e. of v 3 mL (say) of 1(N) HCl (produced)

or

(originally present)

m.e. of Na2 CO3 + m.e. of NaHCO3 = m.e. of v3 mL (say) of 1(N) HCl

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1 1 × 1000 + × 1000 = 1 × v 3 106 84

∴

v3 = 21.3 mL.

Practice Exercise 10. A solution contained Na2CO3 and NaHCO3. 25 mL of this solution required 5 mL of 0.1 N HCl for titration with phenolphthalein as indicator. The titration was repeated with the same volume of the solution but with methyl orange. 12.5 mL of 0.1 N HCl was required this time. Calculate the amount of Na2CO3 and NaHCO3 in the solution.

Answers 10. .053 g, .021 g

1.9 Volume of Strength of H2O2 x volume of H2O2 means x litre of O2 is liberated by 1 litre of H2O2 on decomposition at STP 2H 2 O2  → 2H 2 O + O 2 68 gm

22.4 lit

22.4 lit (at STP) of O2 is given by 68 gm of H2O2 x litre of O2 is released from ∴

x lit of O2 is given by =

Strength,

S=

68 x 17x gm of H2O2 = 22.4 5.6

68 x 17x (strength) = 22.4 5.6

17x 5.6

Normality =

S 17x x [x = N × 5.6] = = E 5.6 × 34 / 2 5.6

Molarity

1 x Normality = . 2 11.2

=

1.10 Strength of Oleum Oleum means mixture of conc. H2SO4 and SO3 Thus, Oleum = x H2SO4 + y SO3 (x H2SO4 + y SO3) + y H2O → (x + y) H2SO4 Strength of oleum is expressed in terms of percentage which is equal to weight of H2SO4 obtained when water is added to 100 gms of oleum. e.g.

109% oleum means 100 gm oleum + water → 109 gm H2SO4 SO3 + H2O  → H2SO4

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9 = Moles of SO3 present in oleum sample. 18

Moles of H2O added = Mass of SO3 in oleum =

9 × 80 = 40 g 18

Thus, oleum sample contained 40% SO3 and 60% H2SO4.

1.11 Oxidation and Reduction Oxidation is (i)

the gain of oxygen

(ii)

the loss of hydrogen

(iii) the loss of electrons (de-electronation) (iv)

the increase of O.N.

Reduction is (i)

the loss of oxygen

(ii)

the gain of hydrogen

(iii) the gain of electron (iv)

the decrease in O.N. Fe

O.N. = 0

 → Fe 2 + + 2e − +2

•

Fe loses electrons

•

There is increase in O.N.

•

Hence Fe is said to be oxidized

•

It is a source of electron hence it can act as a reducing agent (R.A.)

•

Any species that can be oxidized is a reducing agent (R.A.)

•

Cu 2 + + 2e  → Cu O.N.

+2

0

•

Cu2+ gains electron

•

There is decrease in O.N.

•

Hence Cu2+ is said be reduced.

•

Since it can gain electrons, hence it can act as an oxidizing agent (O.A.)

•

Any species that can be reduced is an oxidizing agent (O.A.), or oxidant. Above examples represent half reactions.

A complete reaction showing oxidation and reduction together is called a Redox reaction. Fe + Cu2+

Cu + Fe2+ R O

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Now that you are clear on what is oxidation – reduction, we are now in the right position to know how to balance a redox (oxidation – reduction) reaction. This is important because many of the problems of stoichiometry would be based on such redox eractions.

Oxidation State/Oxidation Number There are several chemical reactions in which oxidation – reduction takes place. Oxidation refers to a reaction in which electrons are removed from an atom and reduction refers to a reaction in which electrons are added to an atom. To describe these changes, the concept of oxidation state becomes necessary. For ionic species, the charge on each ion is said to be the oxidation state for that atom. For example in NaCl, Na exists as Na+ and Cl exists as Cl− . Therefore the oxidation state of Na in NaCl is +1 and that of Cl− is –1. But in covalent molecules, the charge on an atom would be so small that sometimes it becomes impossible to calculate the exact charge on each atom of a molecule. Therefore, the Oxidation State (O.S.) or Oxidation Number (O.N.) is defined as the charge, an atom would have in a molecule if all the bonds associated with this atom in the molecule are considered to be completely ionic. For example in H2O there are two O–H bonds. If we assume both the O–H bonds to be completely ionic, then each H would possess a charge of + 1, while O possess a charge of –2. This is because oxygen is more electronegative than hydrogen. On the other hand, in H2O2 there are two OH bonds and one O–H bond (H–O–O–H). Considering each O–H bond to be ionic both the oxygen atoms acquire a charge of –1 and both the H, +1. This is because O – O bond can not be assumed to be ionic as both the atoms have the same electronegativity. To calculate the oxidation state of an element in a molecule you need not always know the structure of the molecule. There are certain set of rules used to assign oxidation states in polyatomic molecules. (i)

The O.S. of all elements (in any allotropic form) is zero.

(ii)

O.S. of oxygen is –2 in all its compounds except peroxides like H2O2 and Na2O2 where it is –1 and superoxides like KO2 where it is –1/2. In OF2 and O2F2, oxygeneous +2, +1 respectively.

(iii) O.S. of hydrogen is +1 in all of its compounds except those with the metals where it is –1. (iv)

O.S. of all alkali metals is +1 and alkaline earth metals is +2 in all their compounds.

(v)

O.S. of all the halogens is –1 in all their compounds except where they are combined with an element of higher electronegativity. Since fluorine is the most electronegative of all elements, its O.S. is always –1.

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Chapter-1 / Stoichiometry & Redox Reaction

Illustrations Illustration 17 Calculate the O.S. of the atoms in the following species: (i)

In ClO¯ (ii)

(iii) NO3¯

NO2¯

(iv) CCl4

(v)

K2CrO4

(vi) KMnO4

Solution (i)

In ClO¯, the net charge on the species is –1 and therefore the sum of the oxidation states of Cl and O must be equal to –1. Oxygen will have an O.S. of –2 and if the O.S. of Cl is assumed to be ‘x’ then x – 2 should be equal to –1. ∴ x is + 1

(ii)

NO2¯ : (2 × –2) + x = –1 (where ‘x’ is O.S. of N) ∴x=+3

(iii) NO3¯: : x + (3 × –2) = –1 (where ‘x’ is O.s. of N) ∴x+ 5 (iv)

In CCl4, Cl has an O.S. of –1 ∴ x + 4 × –1 = 0 ∴ x = + 4 (where ‘x’ is O.S. of C)

(v)

K2CrO4: K has O.S. of + 1 and O has O.S. of –2 and let Cr has O.S. ‘x’ then, 2 × + 1 + x + 4 × –2 = 0 ∴x=+6

(vi)

KMnO4 : + 1 + x + (4 × –2) = 0 ∴ x = + 7 (where x is O.S. of Mn).

Balancing Redox Equations Some examples of redox reactions are Sn2+ + 2Hg2+

(a)

Hg22+ + Sn4+ R O

MnO4¯ + 5Fe2+ + 8H+ (b)

5Fe3+ + Mn2+ + 4H2O

O R

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6Fe3+ + 2Cr3+ + 7H2O

Cr2O72+ + 6Fe2+ + 14H+ (c)

O R 2NO + 3Cu2+ + 4H2O

3Cu + 2NO3¯ + 8H+

(d)

R O If one of the half reactions does not take place, other half will also not take place. We can say oxidation and reduction go side by side. 5Cl¯ + ClO3¯ + 3H2O

3Cl2 + 6OH¯ O.N. = 0

–1

+5

R

In this we find the Cl2 has been oxidized as well as reduced. Such type of redox reaction is called Disproportionation reaction. Examples are 2Cu+

Disproportionates

Cu + Cu2+

R O 2HCHO + OH¯

CH3OH + HCOO¯ R O

4ClO3¯

3ClO4¯ + Cl¯ O R

3MnO42- + 2H2O

MnO2 + 2MnO4- + 4OH-

R O

How to Balance a Redox Reaction Ion Electron Method This method involves the following steps

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(i)

Chapter-1 / Stoichiometry & Redox Reaction

Divide the complete equation into two half reactions, one representing oxidation and the other reduction.

(ii)

Balance the atoms in each half reaction separately according to the following steps (a)

First of all balance the atoms other than H and O.

(b)

In a reaction taking place in acidic or neutral medium, oxygen atoms are balanced by adding molecules of water to the side deficient in oxygen atoms while hydrogen atoms are balanced by adding H+ ions to the other side deficient in hydrogen atoms. On the other hand, in alkaline medium (OH¯), for every excess of oxygen atom on one side is balanced by adding one H2O to the same side and 2OH¯ to the other side. In case hydrogen is still unbalanced, then balance by adding one OH¯, for every excess of H atom on the same side and one H2O on the other side.

(c)

Equalize the charge on both sides by adding suitable number of electrons to the side deficient in negative charge.

(iii) Multiply the two half reactions by suitable integers so that the total number of electrons gained in one half reaction is equal to the number of electrons lost in the other half reaction. (iv)

Add the two balanced half equations and cancel any term common to both sides.

Illustrations Illustration 18 +

(a)

H NO3− + H 2S  → HSO−4 + NH +4 acidic medium

(b)

OH Fe + N 2 H 4  → Fe(OH)2 + NH3 alkaline medium

−

Solution (a)

Step 1

N5+ + 8e → N3–

Step 2

S2– → S6+ + 8e

Step 3

NO3¯ + H2S → NH4+ + HSO4¯

Step 4

No other atom (except H and O) is unbalanced therefore no need for this step.

Step 5

Balancing O atom. This is made by using H2O and H+ io ns. Add desired molecules of H2O on the side deficient in O atom and double H+ on opposite side. Therefore NO3− + H 2S + H 2 O  → NH 4+ + HSO 4− + 2H +

Step 6

Balancing charge by H+

NO3− + H 2S + H 2 O + 3H +  → NH 4+ + HSO4− + 2H + ∴ Balanced equation is NO3− + H 2S + H 2 O + H +  → NH 4+ + HSO 4− Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail: [email protected]

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(b)

Step 1

Fe → Fe2+ + 2e

Step 2

N22– + 2e → 2N3–

Step 3

Fe + N2H4 → Fe(OH)2 + 2NH3

Step 4

No other atom (except H and O) is unbalanced and therefore no need for this step.

Step 5

Fe + N 2 H 4 + 4OH −  → Fe(OH) 2 + 2NH 3 + 2H 2 O

Step 6

Balance charge by H+ ∴ Fe + N2H4 + 4OH¯ + 4 H+ → Fe(OH)2 + 2NH3 + 2H2O

Finally Fe + N2H4 + 2H2O → Fe(OH)2 + 2NH3

Practice Exercise 11. Balance the redox equation, HNO3 + H2S  → NO + S 12. Balance the following redox equation, FeC2O4 + KMnO4 + H2SO4  → Fe2(SO4)3 + CO2 + MnSO4 + K2SO4.

Answers 11. 2HNO3 + 3H2S  → 3S + 2NO + 4H2O 12. 10FeC2O4 + 6KMnO4 + 24H2SO4  → 5Fe2(SO4)3 + 6MnSO4 + 3K2SO4 + 20CO2 + 24H2O

Oxidation Number Method This method is based on the principle that the number of electrons lost in oxidation must be equal to the number of electrons gained in reduction. The steps to be followed are Write the equation (if it is not complete, then complete it) representing the chemical

(i)

changes. (ii)

By knowing oxidation number of elements, identify which atom(s) is(are) undergoing oxidation and reduction. Write down separate equations for oxidation and reduction.

(iii) Add required electrons on the right hand side of oxidation reaction and on the left hand side of reduction reaction. Care must be taken to ensure that the net charge on both the sides of the equation is same. (iv)

Multiply the oxidation and reduction reactions by suitable numbers to make the number of electrons lost in oxidation reactions equal to the number of electrons gained in reduction reactions.

(v)

Transfer the coefficient of the oxidizing and reducing agents and their products to the main equation. By inspection, arrive at the co-efficient of the species not undergoing oxidation or reduction.

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Chapter-1 / Stoichiometry & Redox Reaction

Illustrations Illustration 19 Balance the equation K2Cr2O7 + HCl → KCl + CrCl3 + H2O + Cl2 Solution +6

−1

+3

0

K 2 Cr2 O7 + HCl  → KCl + CrCl3 + H 2 O + Cl2

Thus here cr of K2Cr2O7 is reduced to CrCl3*(+6 → + 3) and Cl of HCl is oxidized to Cl2 (–1 → 0). In short. Oxidation:

2Cl −  → Cl02 ↑ 2

…(a)

Reduction:

Cr2+6  → 2Cr +3 ↓ 6

…(b)

Step (iii)

2Cl−1  → Cl20 + 2e − ; Cr2+6 + 6e− → 2Cr 3+

Step (iv)

Multiply equation (a) by 6 and (b) by 2 12Cl−1  → 6Cl02 + 12e− 2Cr2+6 + 12e −  → 4Cr +3 2Cr2+6 + 12Cl−1  → 4Cr 3 + 6Cl02

or

Cr2+6 + 6Cl−  → 2Cr +3 + 3Cl02

Step (v)

K2Cr2O7 + 6HCl → 2CrCl3 + 3Cl2

Step (vi)

Making provision of KCl and H2O in the product: Since the reactant has 7 oxygen atoms in the product 7H2O must be present. For accounting 14 hydrogen atoms of water in the product, the reactants must have 12 HCl molecules (the only H containing species). For accounting the 2K atoms and 14 – 12 = 2 additional Cl atoms in the reactant, the product must have 2KCl. Hence the balanced equation is K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 7H2O + 3Cl2

Practice Exercise 13.

Balance the following oxidation-reduction equationm KMnO4 + KCl + H2SO4  → MnSO4 + K2SO4 + H2O + Cl2

14.

Balance the following redox equation K2Cr2O7 + HCl  → KCl + CrCl3 + Cl2 + H2O

Answers 13. 2KMnO4 + 10KCl + 8H2SO4  → 2MnSO4 + 6K2SO4 + 8H2O + 5Cl2 14. K2Cr2O7 + 14HCl  → 2KCl + 2CrCl3 + 3Cl2 + 7H2O

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Miscellaneous Problems Objective Type (* = Problems of Section-I)

*Example 1 8 g of sulphur is burnt to form SO2, which is oxidized by Cl2 water. The solution is then treated with BaCl2 solution. The amount of BaSO4 precipitated would be (a)

1 mole

(b) 0.5 mole

(c) 0.24 mole

(d) 0.25 mole

Solution O2 Cl2 water BaCl2 S → SO 2  → SO 42 −  → BaSO4 + 2Cl −

Moles of sulphur =

8 1 = = Moles of SO2 32 4

∴

Moles of BaSO4 precipiatated =

∴

Ans. (d)

1 4

(applying principle of atom conservation)

*Example 2 The normally of 0.3 M H3PO3 when it undergoes the following reaction, H3 PO3 + 2OH −  → HPO32 − + 2H 2 O would be (a)

0.6 N

(b) 0.15 N

(c) 0.9 N

(d) 0.1 N

Solution Molar ratio of reactants being 1 : 2, so the n-factor ratio has to be 2 : 1, as n-factor of OH ¯ is 1. ∴

n-factor of H3PO3 in the given reaction is 2.

∴

Normally of H3PO3 = 0.3 × 2 = 0.6 N

∴

Ans. (a)

*Example 3 The moles of ammonium sulphate needed to react with one mole of MnO2 in acidic medium in a reaction giving MnSO4 and (NH4)2S2O8 is (a)

2

(b) 3

(c) 1

(d) 1/3

Solution

MnO 2 n=2

+ (NH 4 ) 2 + 2H 2SO 4  → MnSO 4 + (NH 4 ) 2 S2 O8 + 2H 2 O n =1

n =1

Law of equivalence is not applicable

∴

Ans. (c)

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Chapter-1 / Stoichiometry & Redox Reaction

*Example 4 The number of moles of Cr2 O72 − needed to oxidize 0.136 equivalents of N 2 H 5+ by the reaction N 2 H 5+ + Cr2 O 72 −  → N 2 + Cr 3+ + H 2 O is

(a)

0.136

(b) 0.272

(c) 0.816

(d) 0.0227

Solution

 +6  ‘n’ factor of Cr2 O27 − ion in the given reaction is 6.  Cr2 O 72 −  → Cr 3 +    Equivalents of Cr2O72– needed to oxide N 2 H 5+ = Equivalents of N 2 H 5+ =0.136 ∴

Moles of Cr2 O72 − needed to oxides N 2 H 5+ =

∴

Ans. (d)

0.136 = 0.0227 6

*Example 5 A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 g. The % by mass of NaCl in the original mixture is (a)

15.2%

(b) 67.9%

(c) 21.8%

(d) 32.1%

Solution CaCl 2 + Na 2 CO3  → CaCO3 + 2NaCl ∆ CaCO 3  → CaO + CO 2

Moles of CaO =

1.62 = Moles of CaCO3 = Moles of CaCl2 56

Mass of CaCl2 =

1.62 × 111 = 3.21 g 56

% of CaCl2 =

[from equation (i) and (ii)]

3.21 × 100 = 32.1% 10

and % of NaCl = 100 – 32.1 = 67.9 % ∴

Ans. (d)

*Example 6 Equal volumes of 1 M each of KMnO4 and K2Cr2O7 are used to oxidize Fe(II) solution in acidic medium. The amount of Fe oxidized would be (a)

more with KMnO4

(b) equal with both oxidizing agents

(c)

more with K2Cr2O7

(d) cannot be determined

Solution

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The ‘n’ factor of KMnO4 in its reaction with Fe(II) would be 5 while that of K2Cr2O7 would be 6. So, for the same number of moles, K2Cr2O7 will have greater equivalents than KMnO4. So, more equivalents of Fe(II) would be oxidized by K2Cr2O7 than by KMnO4. ∴

Ans. (c)

*Example 7 It takes 2.56 × 10–3 equivalents of KOH to neutralize 0.1254 g H2XO4. The number of neutrons in X would be (a)

16

(b) 8

(c) 18

(d) 32

Solution Let ‘a’ be the atomic mass of X. Moles of H2XO4 =

0.1254 (a + 66)

‘n’ factor of H2XO4 = 2

[as H2XO4 is a dibasic acid]

Equivalents of KOH = Equivalents of H2XO4 ∴

2.56 × 10–3 =

0.1254 ×2 (a + 66)

a = 31.96 g/mol ≈ 32 g/mol ∴

X is sulphur and sulphur has 16 protons and 16 neutrons.

∴

Ans. (a)

*Example 8 10 ml of a solution of H2O2 requires 25 ml of 0.1 N KMnO4 for complete reaction. The volume strength of H2O2 would be (a)

1.4

(b) 1.2

(c) 0.25

(d) 2.5

Solution In the reaction between KMnO4 and H2O2, KMnO4 acts as oxidizing agent, changing to Mn2+ and H2O2 behaves as reducing agent oxidizing to O2.

KMnO 4 + H 2 O 2  → Mn 2 + + O 2 (n = 5)

(n = 2)

Equivalents of KMnO4 = Equivalents of H2O2. 25 × 10–3 × 0.1 = 10 × 10–3 × N N = 0.25

∴

Volume strength of H2O2 = 5.6 × 0.25 = 1.4

∴

Ans. (a)

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Chapter-1 / Stoichiometry & Redox Reaction

Subjective Type Example 1 P and Q are two elements which forms P2Q3 and PQ2. If 0.15 mole of P2Q3 weights 15.9 g and 0.15 mole of PQ2 weights 9.3 g, what are atomic weights of P and Q? Solution Let atomic weight of P and Q a and b respectively ∴

Molecular weight of P2Q3 = 2a + 3b

and Molecular weight of PQ2 = a + 2b Now given that 0.15 mole of P2Q3 weight 15.9 g (2a + 3b) =

  wt. = mole  ∴  mol. wt 

15.9 0.15

Similarly, (a + 2b) =

9.3 0.15

Solving these two equations b = 18 a = 26 Example 2 Potassium selenate is isomorphous with potassium sulphate and contains 45.52% selenium by weight. Calculate the atomic weight of selenium. Also report the equivalent weight of potassium selenate. Solution Potassium seleante is isomorphous to K2SO4 and thus its molecular formula of K2SeO4. Now molecular weight of K2SeO4 = (39 × 2 + a + 4 × 16) where a is atomic weight of Se (142 + a)g K2SeO4 has Se = ag

∴

100 g K2SeO4 has Se =

∵

% of Se = 45.52

∴

a × 100 = 45.52 142 + a

∴

a = 118.168 = 118.2

a × 100 142 + a

Also equivalent of K2SeO4 =

Mol.wt. 2 × 39 + 118.2 + 64 = 130.1. = 2 2

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Example 3 A sample of H2SO4 (density 1.787 g mL–1) is labeled as 86% by weight. What is molarity of acid? What volume of acid has to be used to make 1 litre of 0.2 H2SO4? Solution H2SO4 is 86% be weight ∴

Weight of H2SO4 = 86 g Weight of solution = 100 g

∴

Volume of solution =

∴

M H2SO4 =

100 100 mL = litre 1.787 1.787 × 1000

86 = 15.68 100 98 × 1.787 × 1000

Let V mL of this H2SO4 are used to prepare 1 litre of 0.2 M H2SO4

∴

mM of H2SO4 conc. = mH of H2SO4 dilute

V × 15.68 = 1000 × 0.2 ∴ V = 12.75 mL.

*Example 4 The molecular mass of an organic acid was determined by the study of its barium salt 4.209 g of salt was quantitatively converted to free acid by the reaction with 21.64 ml of 0.477 M H2SO4. The barium salt was found to have two mole of water of hydration per Ba2+ ion and the acid is mono basic. What is molecular weight of anhydrous acid? Solution Meq. of barium salt = meq. of acid

4.290 × 1000 = 21.64 × 0.4777 × 2 M/2 Molecular weight of salt = 415.61 Molecular weight of anion =

∴

415.61 − 137.36 = 121.31 2

Molecular weight of acid = 121.31.

*Example 5 25 mL of a solution of Na2CO3 having a specific gravity of 1.25 g mL–1 required 32.9 mL of a solution of HCl containing 109.5 gm of the acid per litre for complete neutralization. Calculate the volume of 0.84 N H2SO4 that will be completely neutralized by 125 g of Na2CO3 solution. Solution

N HCl =

109.5 =3 36.5 × 1

Since Na2CO3 is completely neutralized by HCl

∴

Meq. of Na2CO3 = Meq. of HCl

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Chapter-1 / Stoichiometry & Redox Reaction

N × 25 = 32.9 × 3 ∴

N Na 2CO3 = 3.948

Now Na2CO3 fresh solution reacts with H2SO4 ∴

Volume of Na2CO3 solution =

∴

Meq. of H2SO4 = Meq. of Na2CO3 0.84 × V = 100 × 3.948

∴

Volume of H2SO4 = 470 mL.

*Example 6 Borax in water gives

B4 O72 − + 7H 2 O  → 4H3 BO3 + 2OH − How many gram of borax (Na2B4O7.10H2O) are required to? (a) Prepare 50 mL of 0.2 M solution (b) neutralize 25 mL of 0.1934 M of HCl and H2SO4 separately Solution

  Normality ∵ Molarity = −  No. of replaceable OH   ∴

N=M×2

Thus Meq. of borax in solution = 50 × 0.2 × 2 = 20 ∴

w × 1000 = 20 M/2

∴

w × 1000 = 20 282 / 2

∴

w = 3.82 g

For neutralization of HCl Meq. of HCl = Meq. of borax 25 × 0.1934 = ∴

w × 1000 382 / 2

Weight of borax = 0.09235 g

For neutralization of H2SO4 Meq. of borax = Meq. of H2SO4

w × 1000 = 25 × 0.1934 × 2 382 / 2 ∴

Weight of borax = 1.847 g.

*Example 7 A mixture containing As2O3 and As2O5 required 20.10 mL of 0.05 N iodine for titration. The resulting solution is then acidified and excess of KI was added. The liberated iodine required 1.1113 g hypo (Na2S2O3 . 5H2O) for complete reaction. Calculate mass mixture. The reactions are Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail: [email protected]

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As2O3 + 2I2 + 2H2O  → As2O5 + 4H+ + 4I¯ As2O5 + 4H+ + 4I¯  → As2O3 + 2I2 + 2H2O Solution Meq. of I2 used = 20.10 × 0.05 = 1.005 Let Meq. of As2O3 Meq. of As2O5 in mixture be a and b respectively. On addition of I2 to mixture. As +2 3 is converted to As +2 5 . ∴

Meq. of As2O5 = Meq. of I2 to mixture used = 1.005 – Meq. of As5+ formed.

or

a = 1.005

...(1)

After the reaction with I2, mixture contains all the arsenic in +5 oxidation state which is then titrated using KI + hypo. Thus, Meq. of As2O3 as As+5 + Meq. of As2O5 as As+5 = Meq. of liberated I2 = Meq. of hypo used

1.113 × 100 248

or

a+b=

or

a + b = 4.481

By equations (1) and (2), B = 4.481 – 1.005 = 3.476 ∴

Wt. of As2O3 =

and. Wt. of As2O5 = ∴

Meq. × Eq.Wt 1.005 × 198 = = 0.0497g 1000 4 × 100 3.476 × 230 = 0.19999g 3 4 × 1000

Wt. of mixture = 0.0497 + 0.1999 = 0.2496 g.

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Chapter-1 / Stoichiometry & Redox Reaction

Exercise - I General Type (Fill in the blanks/ True or False/ Assertion & Reason ) True/False 1.

Mass of 3.01 × 1023 molecules of methane is 8 g.

2.

AgNO3 + NaCl → AgCl + NaNO3 is a substitution reaction.

3.

Exact atomic masses of elements can be determined by applying Dulong and Petit’s law.

4.

The vapour density of ozone is 24

5.

18 g of water vapour and 18 g of ice will not contain the same number of molecules.

Fill in the Blanks 1.

3 g of a salt of molecular weight 30 is dissolved in 250 g of water. The molality of the solution is _____

2.

The weight of 1 x 1022 molecules of CuSO4.5H2O is _____

*3.

The compound YBa2Cu3O7, which shows superconductivity, has copper in oxidation state _____ Assume that the rare earth element yttrium is in its usual + 3 oxidation state.

*4.

In particular concentration of HCN, IO3¯ oxidizes I2 according to the reaction IO3¯ + I2 + CN¯ → 5I (CN)2¯ + 3H2O In this reaction, the molar ratio of IO3¯ : I2 is equal to …….. and the equivalent weight of IO3¯ is molecular wt. divided by ………

*5.

The oxidation number of Cr in CrO5 is ………. .

*6.

The oxidation number of S in H2S2O8 and H2SO5 is ………. and ………. respectively.

*7.

The oxidation number of S and C in Fe(SCN)2 is ………. and ………. respectively

8.

Hydrazine (N2H4) acts as reducing agent and evolves N2 gas when reacts with I2. Then the molar ratio of N2H4 to I2 is ………. and the equivalent wt of N2H4 is ………. .

*9.

When Fe(SCN)2 reacts with Ba(MnO4)2 in acidic solution, then Fe(SCN)2 oxidises into Fe+3, SO4+2, NO3¯ and CO3–2. Then the molar ratio of ratio of Fe(SCN)2 to Ba(MnO4)2 is ………. .

*10.

In a redox reaction, BrO3– (bromate ion) reduces into Br–, then its molecular weight will be equal to equivalent wt. multiplied by ……. .

*11.

When I2 oxidises Na2S2O3 into SO4–2 in basic medium and I2 reduces into I¯ , then the molar ratio of I2 to Na2S2O3 is ………. .

*12.

For the redox reaction MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O, then n-factor of HCl is ………. .

*13.

x M FeS2 has ………. N while it oxidizes into Fe2O3 and SO2 with excess O2.

*14.

In redox process HCN oxidizes into CO3–2 and NO3–, So its n-factor is ………. and M is ………. N respectively.

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Assertion and Reason (a) (b) (c) (d)

If both A and R are true and R is the correct explanation of A. If both A and R are true but R is not the correct explanation of A If A is true but R is false. If A is false but R is true.

*1.

A : In the balanced redox reaction x Cu2O + y NO3– + 14 H+  → 6 Cu+2 + 2NO + 7H2O, the n-factor of Cu2O and NO3– is 2 and 3 respectively. R : since reciprocal of n-factor’s ratio is molar ratio and so, x : y = 3 : 2.

*2.

A : For the redox reaction Cu2S → CuO + SO2, y mole Cu2S in equal to 8y equivalents of Cu2S R : Since the n-factor of Cu2S is 8 and number of equivalents = n–factor × number of moles.

*3.

A : For the reaction Na2CO3 + HCl → NaCl + NaHCO3 the equivalent weight of Na2HCO3 is 106. R : Because the n-factor Na2CO3 is 1 and equivalent weight =

*4.

molecular wt. n − factor

A : In brown ring complex, [Fe(H2O)5NO]SO4, the oxidation number of iron is + 1. R : Because due to charge transfer the one unpaired electron of NO shifts to Fe+2 thereby Fe+2 converts into Fe+.

*5.

A : If the brown ring complex, [Fe(H2O)5NO]SO4 by reacting with HCl converts into FeCl3, NOCl and H2SO4 then the equivalent weight of the complex will be equal to its molecular weight devided by two. R : Because the n-factor of the brown ring complex will be two.

*6.

A : The lowest-oxidation state or oxidation number of N is –3 R : Because the lowest oxidation number of non-metal = V.E. is valence electron.

*7.

A : The number of moles of KMnO4 that will be needed to react with 1 mole of HCN (While HCN oxidizes into C3¯ and NO3¯) in acidic medium is 0.5. R : Because the n-factor of KMnO4 in acidic medium is 5 and the n-factor of HCN is 10. So, molar ratio

*of KMnO4 to HCN is 1 : 2. *8.

A : The oxidation number of “S” in H2S2O8 is 7. R : The maximum oxidation number of S is +6, because the maximum oxidation of element is its no. of valence electron (s) or, group number.

*9.

A : A bottle is labeled as “10 V” of H2O2. so, its percentage strength is 5%. R : % strength of H2O2 is nothing but it is the number of grams of H2O2 in 100 mL solution of H2O2 and it is related with volume strength as under 1 “Vol” of H2O2 = 0.303%.

*10. A :

When a 20 mL of mixture (whose one litre contain x m mole, y m mole and z m mole of NaOH, Na2CO3 and NaHCO3 respectively), is treated with 1 N HCl in presence of methyl orange indicator

2y z   x from the very beginning, then the total used volume of HCl is  + +  ml 100 100 100   R : Because the n-factor of Na2CO3 reacting with HCl is 1 and total m eq’s of HCl used is equal to total m eq’s of bases reacted.

***** -42-

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Chapter-1 / Stoichiometry & Redox Reaction

Exercise - II Objective Type (Only One Choice Correct) Level – I *1.

20 g of a sample of Ba(OH)2 is dissolved in 50 ml of 0.1 NHCl solution. The excess of HCl was titrated with 0.1 N NaOH. The volume of NaOH used was 20 cc. The percentage of Ba(OH)2 in the sample. (a) 128%

*2.

(b) 1.25%

(d) 4.08

It takes 2.56 ×10–3 equivalents of KOH to neutralize 0.125 g H2XO4. The number neutrons in x is (a) 16

*3.

(c) 2.28

(b) 8

(c) 7

(d) 32

0.1 lit of 0.01 M KMnO4 oxidized 100 ml H2O2 in acidic medium. Volume of same KMnO4 required in alkaline medium to oxidize 0.1 lit of same H2O2 will be (Mn changes to Mn+2 in acidic medium and MnO2 in alkaline medium) (a) 100/3 ml

4.

(b) 500/3 ml

(c) 300/5 ml

(d) None

The hydrated salt Na2SO4. nH2O, undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be (a)

*5.

5

(b) 3

(c) 7

(d) 10

8 g of sulphur is burnt to form SO2 which is oxidized by Cl2 water. The solution is treated with BaCl2 solution. The amount of BaSO4 precipitated is (a)

*6.

1 mole

(b) 0.45 mole

(c) 0.24 mole

(d) 0.25 mole

A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate the calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 gms. The % by mass of CaO is 1.62 gms. The % by mass of CaCl2 in the original mixture is (a)

*7.

15.2%

(b) 32.1 %

(d) 11.07%

How many volumes of 1 M KMnO4 ion will react with 1 mol of ferrous oxalate in acidic medium (a) 1/5

*8.

(c) 21.8 %

(b) 2/5

(c) 3/5

(d) 5/3

If equal volumes of 10 M KMnO4 and 10 M K2Cr2O7 solutions are allowed to oxidize Fe(II) to Fe(III) in acidic medium, then Fe(II) oxidized will be

*9.

(a)

more by KMnO4

(b) more by K2Cr2O7

(c)

equal in both cases

(d) can’t be determined

10 ml of a solution of H2O2 required 25 ml of 0.1 N KMnO4 for complete reaction. The volume strength of H2O2 is (a)

10.

1.4

(b) 1.2

(c) 0.25

(d) 2.5

When 10 ml of ethyl alcohol (density = 0.7893 g/mol) is mixed with 20 ml of water (density = 0.9971 g/mol) at 25°C, the final solution has a density of 0.9571 g/ml. The percentage change in total volume on mixing is (a) – 3.1 %

(b) 2.4 %

(c) 1 %

(d) 10%

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11.

Calculate the number of oxygen atoms required to combine with 7 g of N2 to form N2O3 when 80% of N2 is converted into N2O3 (a) 2.3×1023

*12.

(b) 3.6 ×1023

(c) 1.8 ×1021

(d) 5.4 ×1021

One mole of a mixture of CO and CO2 requires exactly 20 grams of NaOH to convert all the CO2 into Na2CO3. How many more grams of NaOH would it require for conversion into Na2CO3 if the mixture (one mole) is completely oxidized to CO2 (a) 60 g

*13.

(b) 80 g

(c) 40 g

(d) 20 g

A solution containing both Na2CO3 and NaHCO3 was treated with excess of CaCl2 solution and filtered. The precipitate weighed M1 gram. On adding NaOH in drop to filtrate avoiding excess, a further M2 grams was precipitated. If after adding excess of CaCl2, the solution (had not be filtered out) was simply boiled and then filtered, what would be the total weight of the precipitate (a) ( M1 + M 2 ) g

14.

 

(b)  M1 +

M2 2

 g 

 M1 + M 2 2 

(c) 

 g 

 

(d)  M 2

M1  g 2 

The density of 1 M solution of NaCl is 1.0585 g/ml. The molality of the solution is (a) 1.0585

15.

(b) 1.00

(c) 0.10

(d) 0.0585

1 mole of Fe2S3, 2 moles of H2O, 3 moles of O2 are allowed to react according to the equation

2Fe 2S3 ( s ) + 6H 2 O ( l ) + 3O 2 ( g ) → 4Fe ( OH )3 ( s ) + 6S ( s ) The number of moles of Fe(OH)3(s) that can be produced is (a)

*16.

1.34

(b) 2

(c) 4

(d) 3

In the mixture of NaHCO3 and Na2CO3, volume of given HCL required is x ml with phenolphthalein indicator and in an another experiment y ml required with methyl orange indicator. Hence volume of HCl for complete reaction of NaHCO3 is (a) 2x

*17.

(b) y

(c) x/2

(d) (y–x)

In a reaction FeS 2 + KMnO 4 + H + → Fe +3 + SO 2 + Mn +2 + H 2 O the equivalent mass of FeS2 would be equal to (a)

*18.

(c) molar mass/11

(d) molar mass/13

40%

(b) 80%

(c) 60%

(d) 9%

An element (X) having equivalent weight E forms a general oxide XmOn, it’s atomic weight should be (a)

*20.

(b) molar mass/10

A sample of oleum is labeled 109%. The % of free SO3 in the sample is (a)

*19.

molar mass

2En / m

(b) 2 m En

(c) E/n

(d) ME/2n

The level on a bottle of H2O2 solution reads as ’10 volume’. The concentration of H2O2 in percentage by volume is (a)

4.05 %

(b) 3.03 %

(c) 6.06%

(d) 2.03 %

Level - II *21. One mole of N2H4 loses ten moles of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (There is no change in the oxidation state of hydrogen). (a) –1

-44-

(b)

–3

(c)

+3

(d) + 5

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Chapter-1 / Stoichiometry & Redox Reaction

22. A molal solution is one that contains one mole of a solute in: (a) 1000 g of the solvent

(b) one litre of the solvent

(c) one litre of the solution

(d) 22.4 litres of the solution

(1986-1 Mark)

*23. The equivalent weight of MnSO4 is half its molecular weight when it is converted to: (a) Mn2O3

(b)

MnO2

(c)

MnO −4

(d)

MnO 24 −

*24. For the redox reaction: MnO −4 + C 2 O −4 + H +  → Mn 2 + + CO2 + H 2 O The correct coefficients of the reactants for the balanced reaction are MnO −4

C 2 O 4−

H+

(a) 2

5

(b) 16

5

2

(c) 5

16

2

(d) 2

16

5

*25. The

16

number of moles of KMnO4 that will be needed to react completely with one mole of ferrous

oxalate in acidic solution is (a)

3 5

(b)

2 5

(c)

4 5

(c)

0.3

(d) 1

*26. The normality of 0.3 M phosphorous acid (H3PO3) is, (a) 0.l

(b)

0.9

(d) 0.6

→ ClO 3− (aq) + 2Cl − (aq), is an example of 27. The reaction, 3ClO− (aq)  (a) oxidation reaction

(b)

reduction reaction

(c) disproportionation reaction

(d) decomposition reaction

*28. An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1N NaOH required to completely neutralize 10 ml of this solution is (a) 40 ml

(b)

20 ml

(c)

10 ml

(d) 4 ml

*29. In the standardization of Na2S2O3 using K2Cr2O7 by iodometry, the equivalent weight of K2Cr2O7 is (a) (molecular weight)/2

(b)

(molecular weight)/6

(c) (molecular weight)/3

(d) same as molecular weight

30. How many moles of electron weigh one kilogram? (b)

1 × 10 31 9.108

(d)

1 × 10 8 9.108 × 6.023

(a) 24g of C (12)

(b)

56g of Fe (56)

(c) 27g of Al (27)

(d) 108g of Ag (108)

(a) 6.023x1023 (c)

6.023 × 10 54 9.108

31. Which has maximum number of atoms?

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Exercise - III Objective Type (More than One Choice Correct) *1.

for the following balanced redox reaction, 2MnO4¯ + 8H+ + Br2 ↽ ⇀ 2Mn+2 + 2 BrO3¯ + 2H2O If the molecular wt. of MnO4¯, Br2 be Mx, My respectively, then

*2.

(a) equivalent wt. of MnO4¯ is Mx/5

(b)

equivalent wt. of Br2 is My/10

(c) the n-factor ratio of Mn+2 to BrO3¯ is 1 : 1

(d) none of these

When a mixture of Cu2S and CuS is titrated with Al(MnO4) in acidic medium, then the oxidation product of Cu2S and CuS are Cu+2, and Al(MnO4)3 be M, M′ and M″ respectively, then which of the following statement are correct? (a) Equivalent weight of Cu2S is

M 8

(c) Equivalent weight of Al(MnO4)3 is M″/5

*3.

(b)

Equivalent weight of CuS is M′/5

(d) Equivalent weight of Al(MnO4)3 is M″/15.

For the balanced redox reaction a NO3¯ + b As2S3 + 4 H2O  → x AsO4– 3 + y NO + z 3 SO4–2 + 8H which of the following statements are correct? (a)

equivalent wt. of As2S3 is M/28 where M is molecular wt. of As2S3

(b) The value of a : b = 28 : 3

*4.

(c) The value of

a + 2b is 1 x+y

(d) The value of

z−x is 1 3

When non-stoichiometric compound Fe0.95O is heated in presence of oxygen, then it converts into Fe2O3, then which of the following statements are correct? (a) Equivalent weight of Fe0.95O is

M where M is molecular weight of Fe0.95O 0.85

(b) The number of moles of Fe+3 and Fe+2 1 moles Fe0.95O and 0.1 and 0.85 respectively (c) The number of moles of Fe+3 and Fe+2 in 1 mole of Fe0.95O are 0.85 and 0.10 respectively (d) The % composition of Fe+2 and Fe+3 in the non stoichiometric compound is 89.47% and 10.53% respectively

*5.

When FeS2 is oxidized with sufficient O2, then its oxidation product is found to be Fe2O3 and SO2, if the molecular weight of FeS2, Fe2O3 and SO2 are M, M′ and M″, then which of the following statements are correct? (a) Equivalent wt. of FeS2 is M/11

(b)

(c) The molar ratio of FeS2 to O2 is 11 : 4

(d) The molar ratio of Fe2O3 and SO2 is 1 : 4

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The molar ratio of FeS2 to O2 is 4 : 11

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*6.

Chapter-1 / Stoichiometry & Redox Reaction

40 g NaOH, 106 g Na2CO3 and 84 g NaHCO3 is dissolved in water and the solution is made 1 litre. 20 mL of this stock solution is titrated with 1N HCl, hence which of the following statements are correct? (a)

the titre reading of HCl will be 40 mL, if phenolphthalein is used indicator from the very beginning

(b) the titre reading of HCl will be 60 mL if phenolphthalein is used indicator from the very beginning. (c) the titre reading of HCl will be 40 mL if the methyl orange is used indicator after the 1st end point (d) the titre reading of HCl will be 880 mL, if methyl orange is used as indicator from the very beginning

*7.

150 mL

M Ba(MnO4)2 in acidic can oxidize completely 10

(a) 150 mL 1M Fe+2

(b)

(c) 75 mL 1 M C2O

(d) 25 mL 1M K2Cr2O7 solution

4–2

8.

Which of the following quantities are dependent on temperature? (a) Molarity

*9.

50 mL 1M FeCrO4

COOH

(b)

COOK and

COOH

Normality

(c)

Molality

(d) Mole fraction

behaves as acid are well as reducing agent. Then which of the following

COOH

are the correct statements regarding

(a) When

COOH

COOK and

COOH

behaves as reducing agent, then its equivalent weighted are

COOH

equal to half of its molecular weight respectively? (b) 1000 mL of 1 N solution of each is neutralized by 1000 mL of 1 N Ca(OH)2 (c) 1000 mL of 1 N solution is neutralized by 1000 mL of 1 M Ca (OH)2 (d) 1000 mL of 1 M solution is neutralized by 20 mL 2 of KMnO4 in acidic medium

*10. For the reaction H 3 PO 4 + Ca(OH)2  → CaHPO 4 + 2H 2 O 1 mole

1 mole

Then which of the following statements are correct? (a) the equivalent weight of H3PO4 is 49 (b) the resulting solution is neutralized by 1 mole of KOH (c) 1 mole of H3PO4 is completely neutralized by 1.5 mole of Ca(OH)2 (d) None 11.

The density of 3M sodium thiosulphate is 1.25 g mL–1. Identify the correct statements among the following

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(a)

% by the weight of sodium thiosulphate is 37.92

(b)

The mole fraction of sodium thiosulphate is 0.065

(c)

The molality of Na+ is 2.53 and S2O32– is 1.25

(d)

All of these

A 5 L vessel contains 2.8 g of N2. When heated to 1800 K, 30% molecules are dissociated into atoms

12.

(a)

Total number of moles in the container will be 0.13

(b)

Total number of molecules in the container will be close to 0.421 × 1023

(c)

Total number of moles in the container will be 0.098

(d)

All of these

The density of air is 0.001293 g/cm3 at STP. Identify which of the following statement is correct

13.

*14.

15.

(a)

Vapour density is 14.48

(b)

Molecular weight is 28.96

(c)

Vapour density is 0.001293 g / cm3

(d)

Vapour density and molecular weight cannot be determined

Cr2O72– is reduced to Cr3+ by Fe2+. Identify the incorrect statement from the following (a)

6 moles of Fe2+ are oxidised to Fe3+ ions

(b) The solution becomes green

(c)

The solution becomes yellow

(d) 3 moles of Fe2+ get oxidised to Fe3+

Which of the following has same mass (a)

1.0 mole of O2

(b) 3 × 1023 molecules of SO2

(c)

0.5 moles of CO2

(b) 1 g atom of sulphur

A solution of KMnO4 to be used in acidic medium is prepared by dissolving 1.58 g L–1. The solution

16.

is (a) 17.

M/100

(b) N/20

(c) M/50

(d) N/50

The mole fraction of NaCl in aqueous solution is 0.2. The solution is

18.

(a)

13.9 m

(b) Mole fraction of H2O is 0.8

(c)

Acidic in nature

(d) Neutral

100 mL of 0.06 M Ca(NO3)2 is added to 50 mL of 0.06 M Na2C2O4. After the reaction is complete

*19.

(a)

0.003 moles of calcium oxalate will get precipitated

(b)

0.003 M of excess of Ca2+ will remain in excess

(c)

Na2C2O4 is limiting reagent

(d)

Ca(NO3)2 is excess reagent

0.1 M HCl is used to react with 1g of mixture of Na2CO3 and NaHCO3 containing equimolar amount of two. The volume of HCl used is (a)

20.

156.0 mL

(b) 0.156 L

(c) 1.57 L

(d) 15.7 L

(b) Tribasic acid

(c) Dibasic acid

(d) Aprotic acid

H3BO3, Boric acid is (a)

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Monobasic acid

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*21.

Chapter-1 / Stoichiometry & Redox Reaction

In the titration of K2Cr2O7 and ferrous sulphate, following data is obtained V1 mL of 1.0 M1 K2Cr2O7 requires V2 mL of 1.0 M2FeSO4 Which of the following relations is/are true for the above titration (a)

*22.

*23.

6 M1V2 = M2V2

(b) M1V1 = 6 M2V2

(c) N1V1 = N2V2

(d) M1V1 = M2V2

10 volume H2O2 solution is present, then it means (a)

10 ml of H2O2 solution liberates 1 ml of oxygen at STP

(b)

1 ml of H2O2 solution liberates 10 ml of oxygen at STP

(c)

0.0303 g of H2O2 in 10 ml of solution liberates 10 ml O2 at STP

(d)

0.0303 g of H2O2 in 1 ml of the solution liberates 10 ml O2 at STP

Which of the following statements are correct (a)

0.2 moles of KMnO4 will oxidise one mole of ferrous ions to ferric ions in acidic medium

(b)

1.5 moles of KMnO4 will oxidise 1 mole of ferrous oxalate to ferric oxalate in acidic medium

(c)

0.6 moles of KMnO4 will oxidise 1 mole of ferrous oxalate to one mole of ferric ion and carbon dioxide in acidic medium

(d)

1 mole of K2Cr2O7 will oxidise 2 moles of ferrous oxalate to ferric ions and carbon dioxide in acidic medium

*24.

*25.

Silver metal in ore is dissolved by potassium cyanide solution in the presence of air by the reaction (a)

The amount of KCN required to dissolve 100 g of pure Ag is 120 g

(b)

The amount of oxygen used in this process is 0.742 g

(c)

The amount of oxygen used in this process is 7.40 g

(d)

The volume of oxygen used at STP is 5.20 litres

16 ml of hydride of nitrogen were exploded in a eudiometric, tube and the volume after explosion was 32 ml. To this, 14 ml of oxygen were added and the mixture again was exploded. On cooling, the volume of the residual gases was found to be 100 mL (a)

The volume of the mixture before second explosion is 50 mL

(b)

The volume of the mixture before second explosion is 46 mL

(c)

The residual gases measuring 10 mL is a mixture of N2+ unreacted oxygen

(d)

The decrease in volume after second explosion is 36 mL

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IIT/TYP/U1/Chemistry/07

Exercise - IV Comprehensive/Matching Type *Passage – 1 Iodine titrations: Compound containing iodine are widely used in titrations, commonly known as iodine titration. It is of two kinds (i) Iodometric titrations (ii) Iodimetric titrations (i) Iodometric titrations: It is nothing but an indirect method of estimating the iodine. In this type of titration, an oxidizing agent is made to react with excess of KI, in acidic medium or, basic medium in which I − oxidizes into I2. Now the liberated I2 can be titrated with Na2S2O3 solution. +

Oxidi sin g agent Na 2 S 2 O 3 / H KI  → I 2  → I − + Na 2 S 4 O 6

Although solid I2 is black and insoluble in water, but it converts into soluble I3 ions

I 2 (s) + I − ↽ ⇀ I −3 black dark brown Starch is used as indicator near the end point or equivalence point. Even small amount of I2 molecules, gives blue colour with starch. The completion of the reaction can be detected when blue colour disappears at the and point. In iodimetric titration, the strength of reducing agent is determined by reacting it with I2. 1.

When 79.75 g of CuSO4 sample containing inert impurity is reacted with KI, the liberated I2 is reacted with 50 mL (1M) Na2S2O3 in basic medium, where it oxidizes into SO4–2 ions, and I2 reduces into I − , then what will be the % purity of CuSO4 in sample? (a) 60%

2.

(b)

80%

(c)

50%

(d) 95%

When 214 g of KIO3 reacts with excess of KI in presence of H+, then it reduce I2. Now I2 is completely reacted with 1M Na2S2O3 solution in basic medium, where it converts into SO4–2 ions. Then what volume of Na2S2O3 is needed to react the end point of the reaction? (a) 500 mL

3.

(b)

800 mL

(c)

150 mL

A solution containing Cu+2 and C2O4–2 ions which on titration with

(d) 750 mL

M KMnO4 requires 50 mL. The 10

resulting solution is neutralized with K2CO3, then treated with excess of KI. The liberated I2 required 2 mL

M Na2S2O3 in acidic solution, then what is the difference of the number of m mole of Cu+2 and 10

C2O4–2 ions in the solution? (a) 40 4.

(b)

10

(c)

30

(d) 50

When 1.66 g of KI is reacted with excess of KIO3 in presence of 6N HCl, then ICl is produced. The amount of KIO3 reacted and ICl formed are respectively. (a) 4 × 10–2 mole, 3 × 10–3 mole

(b)

(c) 5 × 10–2 mole, 1.5 × 10–3 mole

(d) 5 × 10–3 mole, 1.5 × 10–2 mole

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1.5 × 10–2 mole, 5 × 10–3 mole

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5.

Chapter-1 / Stoichiometry & Redox Reaction

In order to estimate O3 in the air, 22.4 L of air at NTP, is passed through acidified KI solution where O2 M is evolved and I − is oxidized to I2. The liberated I2 required 200 mL Na2S2O3 in acidic medium, then 10 water will be the % composition of O3 by volume in the sample? (a) 50%

(b)

10%

(c)

0.1%

(d) 1%

*Passage – 2 H2O2 acts as both oxidizing as well as reducing agent. As oxidizing agent, its product is H2O, but as reducing agent, its product is O2. The strength of H2O2 is expressed in many ways like molarity, normality, % strength and volume strength. But out of all these form of strengths, volume strength has great significance for chemical reactions. The decomposition of H2O2 is shown as under: H2O2(l)  → H2O (l) + ½ O2(g) x “vol” strength of H2O2 means 1 volume (litre or mL) of H2O2 releases x volume (litre or mL) of O2 at NTP. 1 litre H2O2 releases x litre of O2 at NTP x = mole of O2 22.4 From equation it is very clear that 1 mole of O2 produces from 2 moles of H2O2. x moles of O2 produces from moles 22.4 x = 2× moles of H2O2 22.4 x = moles of H2O2 11.2 x Since moles of H2O2 is present in 1 litre of H2O2 11.2 x ∴ molarity of H2O2 = 11.2

∴ Normality = n-factor × Molarity x x ∴ Normality of H2O2 = 2 × = 11.2 5.6 Percentage strength of H2O2 is nothing but, it is the weight of H2O2 in gramme per 100 mL of H2O2 solution. By volume strength of H2O2, we can predict the percentage purity of the substance and many more information can be known. 1.

What is the percentage strength of 10 “vol” H2O2? (a) 11%

2.

(b)

5%

(c)

7%

(d) 3.03%

55 g Ba(MnO4)2 sample containing inert impurity is completely reacting with 100 mL of 56 “vol” strength of H2O2 , then what will be the percentage purity of Ba(MnO4)2 in the sample? (Ba–137, Mn– 55) (a) 40%

3.

(b)

25%

(c)

10%

(d) 68.18%

When 100 mL solution of H2O2 reacts completely with KI, then it oxidizes KI into I2. Now liberated iodine required 25 mL of 1 M Na2S2O3 in basic medium, where it oxidizes into SO4–2, then what is the volume strength of H2O2? (a) 5.6

(b)

8.6

(c)

11.2

(d) None

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4.

A 100 mL H2O2 liberates 25.4 g of I2 from an acidified KI solution, then what is the volume strength of H2O2? (a) 5

5.

(b)

(c

10

(d) 11.2

What volume of H2O2 solution of 22.4 “vol” strength is required to liberate 2240 mL of O2 at NTP ? (a) 300 mL

6.

5.6

(b)

200 mL

(c)

500 mL

(d) 100 mL

A 840 mL sample of H2O2 solution containing y “vol” strength of H2O2 requires y mL of Al(MnO4)3 in acidic medium, then what will be the molarity of Al(MnO4)3 solution? (a) 7 M

7.

(b)

5M

(c)

10 M

(d) 15 M

100 mL of a H2O2 solution were diluted to 400 mL, and 20 mL of this diluted solution required to reduce

M 10 mL   KMnO4 solution then what is the “volume” strength of H2O2?  10  (a) 20

(b)

11.2

(c)

8.4

(d) 5.6

*Passage – 3 Normality is number of gram equivalents dissolved per litre of solution. It changes with change in temperature. In case of monobasic acid, normality and molarity are equal but in case of dibasic acid, normality is twice the molarity. In neutralization and redox reaction reactions, number of miliequivalents of reactants as well as products are always equal 1.

On heating a litre of a N/2 HCl solution, 2.750 g of HCl is lost and the volume of solution becomes 750 mL. The normality of resulting solution will be (a)

2.

(d) 5.7

10 mL

(b) 20 mL

(c) 5 mL

(d) 40mL

0.25

(b) 1.0

(c) 0.5

(d) 0.125

6.90 N KOH solution in water contains 30% by weight of KOH. The density of solution will be (a)

5.

(c) 0.057

Molarity of 0.25 N Na2CO3 is (a)

4.

(b) 0.75

The volume of 0.1 M Ca(OH)2 required to neutralize 10 mL of 0.1 N HCl will be (a)

3.

0.58

1.288

(b) 2.88

(c) 0.1288

(d) 12.88

The amount of ferrous ammonium sulphate required to prepare 250 mL of 0.1 N solution is (a)

1.96 g

(b) 1.8 g

(c) 9.8 g

(d) 0.196 g

*Passage – 4 1.00 g of a mixture consisting of equal number of moles of carbonates of two alkali metals, required 44.4 ml of 0.5 N hydrochloric acid for complete reaction. If the atomic weight of one of the metals is 7.00. 1.

The number of equivalents of HCl used is (a)

2.

0.222

(b) 2.22

(c) 22.22

(d) 0.0222

(c) 0.0055

(d) 0.00275

The number of moles of each metal carbonate are (a)

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0.1

(b) 0.0111

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3.

The molecular weights of the two alkali metal carbonate are (a)

4.

Chapter-1 / Stoichiometry & Redox Reaction

74 and 60

(b) 134 and 160

(c) 160 and 60

(d) 74 and 106

The total amount of the sulphate formed on a quantitative conversion of 1.00 g of a mixture into sulphate is (a)

5.

110 g

(b) 1.10 g

(c) 1.4 g

(d) 0.110 g

One gram of a mixture contains (a)

0.022 moles of each metal carbonate

(b)

The two alkali metals are Li and Na respectively

(c)

The two alkali metals are Li and K respectively

(d)

The two alkali metals are Li and Ag

Matching Type Questions 1.

*2.

Column I

Column II

(a)

1 mole of Na

(p) 6.02 × 1023

(b)

1 mole of H2O

(q)

At. wt. in g

(c)

1 mole

(r)

molecular wt. in g

(d) No. of molecules in 16 g CH4

(s)

Avogadro’s number

(a)

(a-p, q, s), (b-q, r, s), (c-p, s), (d-p, s)

(b)

(a-p, s), (b-q, r, s), (c-p, s), (d-p, s)

(c)

(a-p, q, s), (b-q, r, s), (c-p, r, s), (d-p, s)

(d) (a-q, s), (b-q, r, s), (c-p, s), (d-p, s)

Column I

Column II

(a)

Eq. wt. of reducing agent

(p) Formula wt./No. of electron gained per mole

(b)

Eq. wt. of oxidizing agent

(q)

Formula wt./Total change in oxidation No.

(c)

Eq. wt. of salt

(r)

Formula wt./Negative charge

(d) Eq. wt. of salt

(s)

Half the molecular weight

(a)

(a-p, q, s), (b-q, r, s), (c-p, s), (d-p, s)

(b)

(a-p, s), (b-q, r, s), (c-p, s), (d-p, s)

(c)

(a-p, q), (b-q), (c-r), (d-p, q, s)

(d) (a-q, s), (b-q, r, s), (c-p, s), (d-p, s)

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Exercise - V Subjective Type *1.

A 1.00gm sample of H2O2 solution containing X percent H2O2 by weight requires X ml of a KMnO4 solution for complete oxidation under acidic conditions. Calculate the normality of the KMnO4 solution.

*2.

4.08 g of a mixture of BaO and an unknown carbonate MCO3 was heated strongly. The residue weighed 3.64g. This was dissolved in 100 ml of I N HCI. The excess acid required 16 ml of 2.5 N NaOH solutions for complete neutralization. Identify the metal M. (Atomic weight H = 1, C = 12, O = 16, Cl = 35.5, Ba = 138).

*3.

2.68 × 10–3 moles of a solution containing an ion An+ required 1.61 x10–3 moles of MnO−4 for the oxidation of An+ to AO −3 in acid medium. What is the value of n?

*4.

An equal volume of a reducing agent is titrated separately with 1 M KMnO4 in acid neutral and alkaline media. The volumes of KMnO4 required are 20 ml. in acid, 33.4 ml. neutral and 100 ml. in alkaline media. Find out the oxidation state of manganese in each reduction product. Give the balanced equations for all the three half reactions. Find out the volume of 1 M K2Cr2O7 consumed; if the same volume of the reducing agent is titrated in acid medium.

5.

Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is 1.84 g/mI.

*6.

A 2.0 g sample of a mixture of Na 2 CO 3 , NaHCO 3 and Na 2 SO 4 is gently heated till the evolution of

CO 2 ceases. The volume of CO2 at 750 mm of Hg and 298 K is measured to be 123.9 mL. A 1.5 g of the same sample requires 150 mL of (M/10) HCl for complete neutralization. Calculate the % composition of the components of the mixture.

*7.

An aqueous solution containing 0.10 g KIO 3 (formula wt. = 214.0) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated I 2 consumed 45.0 mL of thiosulphate solution to decolorize the blue starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution.

*8.

How many milliliters of 0.5 M H 2 SO 4 are needed to dissolve 0.5 g of copper (II) carbonate?

*9.

Hydrogen peroxide solution (20 ml) reacts quantitatively with a solution of KMnO 4 (20 mL) acidified with dilute H2 SO 4 . The same volume of the KMnO4 solution is just decolourised by 10 ml of MnSO4 in neutral medium simultaneously forming a dark brown precipitate of hydrated MnO2 . The brown precipitate is dissolved in 10 mL of 0.2 M sodium oxalate under boiling condition in the presence of dilute H 2 SO 4 . Write the balanced equations involved in the reactions and calculate the molarity of H2 O2.

10.

Calculate the molarity of water if its density is 1000 kg/m3.

*****

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Chapter-1 / Stoichiometry & Redox Reaction

Exercise - VI IIT-JEE Problems *1.

*2.

Balance the following equations. (i) Cu 2 O + H + + NO 3−  → Cu 2 + + NO + H 2 O (ii)

K 4 Fe ( CN )6  + H 2 SO 4 + H 2 O  → K 2 SO 4 + FeSO 4 + (NH 4 )2 SO 4 + CO

(iii)

C2 H 5 OH + I 2 + OH −  → CHI 3 + HCO 3− + I − + H2 O

Hydroxylamine reduces iron (Ill) according to the equation: 2NH 2 OH + 4Fe 3 +  → N 2 O(g) ↑ +H 2 O + 4Fe 2 + + 4H + Iron (II) thus produced is estimated by titration with a standard permanganate solution. The reaction is: MnO−4 + 5Fe 2 + + 8H +  → Mn 2 + + 5Fe 3 + + 4H 2 O A 10 ml. sample of hydroxylamine solution was diluted to 1 litre. 50 ml. of this diluted solution was boiled with an excess of iron (Ill) solution. The resulting solution required 12 ml. of 0.02 M KMnO4 solution for complete oxidation of iron (II). Calculate the weight of hydroxylamine in one litre of the original solution. (H = 1, N = 14, O = 16, K = 39, Mn = 55, Fe 56).

*3.

The density of a 3 M sodium thiosulphate solution (Na2S2O3) is 1.25 g per ml. Calculate (i) the percentage by weight of sodium thiosulphate, (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of Na+ and S 2 O 3−2 ions.

*4.

Complete and balance the following reactions: (i) Zn + NO −3  → Zn 2 + + NH 4+ (ii)

Cr2 O72 − + C 2 H 4 O  → C 2 H 4 O 2 + Cr 3 +

(iii)

HNO 3 + HCl  → NO + Cl 2

(iv)

Ce 3 + + S 2 O82 −  → SO42 − + Ce 4 +

(v)

Cl 2 + OH −  → Cl − + ClO−

(vi)

Mn 2 + + PbO 2  → MnO −4 + H 2 O

(vii)

S + OH −  → S 2 − + S 2 O 32 −

(viii) ClO−3 + I − + H 2 SO 4  → Cl − + HSO 4− (ix)

*5.

Ag + + AsH 3  → H 3 AsO 3 + H +

Five ml of 8N nitric acid, 4.8 ml of 5N hydrochloric acid and a certain volume of 17M sulphuric acid are mixed together and made upto 2 litre. Thirty ml. of this acid mixture exactly neutralise 42.9 ml of sodium carbonate solution containing one gram of Na2CO3.10H2O in 100 ml. of water. Calculate the amount in gram of the sulphate ions in solution.

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*6.

*7.

(i)

What is the weight of sodium bromate and molarity of solution necessary to prepare 85.5 ml of 0.672 N solution when the half-cell reaction is BrO −3 + 6H + + 6e −  → Br − + 3H 2 O

(ii)

What would be the weight as well as molarity if the half-cell reaction is? 2BrO −3 + 12H + + 10e −  → Br2 + 6H 2 O

A sample of hydrazine sulphate (N2H6SO4) was dissolved in 100 ml. of water, 10 ml of this solution was reacted with excess of ferric chloride solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required 20 ml. of M/50 potassium permanganate solution. Estimate the amount of hydrazine sulphate in one litre of the solution. Reaction: 4Fe 3 + + N 2 H 4  → N 2 + 4Fe 2 + + 4H +

MnO −4 + 5Fe 2 + + 8H +  → Mn 2 + + 5Fe 3 + + 4H 2 O

*8.

A mixture of H2 C2O 4 (oxalic acid) and NaHC 2 O 4 weighing 2.02 g was dissolved in water and the solution made upto one litre. Ten milliliters of the solution required 3.0 ml of 0.1 N sodium hydroxide solution for complete neutralization. In another experiment, 10.0 ml of the same solution, in hot dilute sulphuric acid medium, required 4.0 ml of 0.1 N potassium permanganate solutions for complete reaction. Calculate the amount of H2 C2O 4 and NaHC2 O 4 in the mixture.

*9.

A solid mixture (5.0 g) consisting of lead nitrate and sodium nitrate was heated below 600°C until the weight of the residue was constant. If the loss in weight is 28.0 percent, find the amount of lead nitrate and sodium nitrate in the mixture.

*10.

A solution of 0.2 g of a compound containing Cu 2 + and C 2 O 42 − ions on titration with 0.02 M KMnO 4 in presence of H2 SO 4 consumes 22.6 mL of the oxidant. The resultant solution is neutralized with Na 2 CO 3 , acidified with dilute acetic acid and treated with excess KI. The liberated iodine requires

11.3 mL of 0.050 M Na 2 S 2 O3 solution for complete reduction. Find out the molar ratio of Cu 2 + and C 2 O 42 − in the compound. Write down the balanced redox reactions involved in the above titrations.

*11.

A 1 g sample of Fe 2 O 3 solid of 55.2 per cent purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made upto 100.0 ml. An aliquot of 25.0 ml of this solution requires 17.0 ml of 0.067 M solution of an oxidant for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration.

*12.

Upon mixing 45.0 ml of 0.25 M lead nitrate solution with 25 mL of 0.10 M chromic sulphate, precipitation of lead sulphate takes. How many mol of lead sulphate are formed? Also, calculate the molar concentration of the species left behind in final solution. Assume that lead sulphate is completely insoluble.

*****

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Chapter-1 / Stoichiometry & Redox Reaction

Answers Exercise - I True/False 1. True

2.

False

3.

False

4.

True

5.

False

Fill in the Blanks

7 3

1.

0.4

2.

4.14 g

3.

+

5. 8. 11.

+6 1 : 2, 16 4:1 y 10, 10

6. 9. 12.

6,6 10 : 33 ½

7. 10. 13.

0, +2 6 11x

14.

Assertion and Reason 1. (a) 2. 6. (a) 7.

(a) (a)

3. 8.

(a) (d)

4.

1 : 2, 4

4. 9.

(a) (d)

5. 10.

(a) (a)

Exercise - II Only One Option is correct Level - I 1. (b) 2. (a) 6. (b) 7. (c) 11. (b) 12. (a) 16. (d) 17. (c) Level 21. 26. 31.

- II (c) (d) (a)

22. 27.

(a) (c)

3. 8. 13. 18.

(b) (b) (b) (c)

4. 9. 14. 19.

(d) (a) (b) (a)

5. 10. 15. 20.

(d) (a) (b) (b)

23. 28.

(b) (a)

24. 29.

(a) (b)

25. 30.

(a) (d)

4. 9. 14. 19. 24.

(a,b,d) (a,b,d) (c,d) (a,b) (a,c)

5. 10. 15. 20. 25.

(a,b) (a,b,c) (a,b,d) (a,d) (b,c,d)

Exercise - III More Than One Choice Correct 1. (a, b, c) 2. (a, d) 6. (a,c) 7. (a,b,c,d) 11. (a,b) 12. (a,b) 16. (a,b) 17. (a,b,d) 21. (a,c) 22. (b,d)

3. 8. 13. 18. 23.

(a,b,c,d) (a,b) (a,b) (a,c,d) (a,c,d)

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Exercise - IV Passage – 1 1. (b)

2.

(d)

3.

(b)

4.

(d)

5.

(d)

Passage – 2 1. (d) 6. (d)

2. 7.

(d) (d)

3.

(c)

4.

(d)

5.

(c)

Passage – 3 1. (a)

2.

(c)

3.

(a)

4.

(a)

5.

(c)

Passage – 4 1. (d)

2.

(c)

3.

(d)

4.

(c)

5.

(b)

Matching Type Questions 1. (a) 2. (c)

Exercise - V Subjective Type 1. 0.588 N 3. 2 5. 10.43 7. 0.062 9. 0.1

2. 4. 6. 8. 10.

Ca +2, +4, +6, 16.66 ml Na2CO3 – 26.5%, Na2SO4 – 31.5% 8.097 ml 55.55 M

Exercise - VI IIT-JEE Level Problem

2. 5. 7. 9. 11. 12.

3. (i) 37.92; (ii) 0.065; (iii) 7.74, 3.87 39.6 g l–1 6.3648 g 6. (i) 1.446 gm, 0.112 M; (ii) 1.7532 g, 0.1344 M 6.5 g 8. 1.12 g, 0.90 g 3.324 g, 1.676 g 10. 1 : 2 6 0.0075, Pb2+ = 0.05357 M, NO −3 = 0.3214 M, Cr 3 + = 0.0714 M

*****

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