# Jee 2014 Booklet1 Hwt Solutions Stoichiometry

August 28, 2017 | Author: varunkohliin | Category: Mole (Unit), Stoichiometry, Molecules, Molar Concentration, Molecular Mass

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Jee 2014 Booklet1 Hwt Solutions Stoichiometry...

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Vidyamandir Classes

Solutions to Home Work Test/Chemistry Basic Stoichiometry 2.

No. of atoms =

3.

Mass =

4.

HWT - 1

0.635  6.023  1023  6.023  1021 63.5

112  16  0.08g 22400 Mass = 0.0018  1  0.0018g .

No. of water molecules = moles  NA

0.0018  6.023  1023  6.023  1019 18 Molar mass is mass of one mole of a substance. 19700  6.023  1023  6.023  1025 Atoms of gold = Moles of gold  N A  197 =

5. 7. 8.

Initial molecules of CO2 = moles of CO2  N A 

200  103  6.023  1023 44

 2.7  1021 Molecules leftover = (2.7  1021 )  (1  1021 )  1.7  1021 Moles leftover =

9.

No. of atoms  No. of moles 

1.7  1021 6  1023 20  103

6.6  1023 3  1026 6  1023

 2.9  103  3  1026

 500

Basic Stoichiometry 1.

Molarity = Moles of H 2SO4 in 1L of solution =

 2.

4.9  1000  98

HWT - 2 Mass of H 2SO4 in 1L of solutions 98

20 100  10M  10 moLdm 3

2HX  Mg   MgX 2  H 2 Let M be equivalent weight of acid. Then M  1 is atomic mass of X. Moles of HX consumed = 1/M and moles of MgX2 produced =

1.301 . According to balanced chemical equation. 24  2(M  1)

1 2  1.301 1.301 1.301    M 24  2(M  1) 12  (M  1) M  11

M = 11 = 1.301 M or M = 36.6

4.

 90   8   2   200     199     202   2004 . Average atomic mass =  100 100 100       Mass of water  NA No. of water molecules = No. of moles of water  N A  18 1000  1   N A  55.55N A 18

5.

 98   2   12     14   12 Average atomic mass =  100 100    

3.

VMC/Basic Stoichiometry

13

HWT-Solutions/Chemistry

Vidyamandir Classes  And

12  6.023  1023  6.023  1023 12 2  6.023  1023 Total No. of C -14 atoms in 12g of C = 100 Total No. of C atoms in 12g of C 

= 12.046  1021  1.20  1022 6.

 7.

(100  x) x  (z  2)   (z  1)  z 100 100 (100  x) ) (z  2)  x (z  1)  100z  x  66.6

Let x% be abundance of lighter one. Then

Let x be atomic mass of metal, M 28  (3x  28)  28  Now, 100

x  24

8.

BaCl 2

 

H 2SO4

Initially

0.1 moles

0.05 moles

Finally

0.05

-

BaSO4

2HCl

-

-

0.05

0.1

Mass of BaSO4  0.05  233  11.65g 9.

CaCO3 

2HCl

Initially

0.2 moles

0.55 moles

Finally

-

0.15 mles

 

CaCl2

 CO2

0.2 moles

H 2O

-

-

0.2 moles

0.2 moles

Mass of CO2 produced = 0.2  44  8.8g 10.

1 moles of NH 3 neutralizes one mole of HCl

292 8 36.5 Moles of NH 3 required = 8

Mass of (NH 4 )SO4 required = 4  132  528 g

Now, moles of HCl 

Moles of (NH 4 )SO4 required = 4

Basic Stoichiometry 1.

Molecules of H 2SO4 present in 1 mole of H 2SO4  6.02  1023

 2. 4.

5.

6.

HWT - 3

Atoms present in 1 mole of H 2SO4  7  6.02  1023 .

Atomic mass of H  1u  1.6  1027 kg  1.6  1024 g Meq. Of metal = Meq. of oxygen 0.1 46.6 1000    1000  2  2  E = 12 E 22400 Atomic weight  Atomic Weight  74.4 Equivalent weight = 37.2 = 2 Molecular weight of chloride = 74.4 + 71 = 145.4 Let x be atomic mass of M. 53  (2x  48) Then 2x  100

VMC/Basic Stoichiometry

 x  27

14

HWT-Solutions/Chemistry

Vidyamandir Classes 7.

Consider 1 L (1000mL) of solution 29  d  1000 where d is density in g/mL Mass of H 2SO4 in solution  100 Also, Mass of H 2SO4 in solution = 3.60  98

 9.

10.

3.60  98 

29  d  1000 100

 d  1.22

1 mole of Mg3 (PO4 )2 contains 8 moles of oxygen atoms.

1 mole of oxygen atoms is contained in 1/8 moles of Mg3 (PO4 )2

0.25 moles of oxygen atoms are contained in 1/32 moles of Mg3 (PO4 )2

1 mole of CO2 contains 6.02  1023 molecules of CO2

1 mole of CO2 contains 6.02  1023 atoms of C.

Basic Stoichiometry 1.

HWT - 4

Mass of 1 mole of e s = Mass of one e   N A  9.1  1031  6.02  1023 kg = 5.46  108 kg  0.55mg

4.

CaCO3 

 

2HCl

Initially

1 mole

1 mole

Finally

0.5 moles

-

 CO2

CaCl2 -

-

0.5 moles

0.5 moles

H 2O 0.5 moles

Weight of liberated CO2  0.5  44  22g 6.

1 gram atom  6.02  1023 atoms

 7.

1 atom 

1 6.02  10

23

 1.66  1024 gm atoms.

222 2 111 No. of CaCl 2 molecules = 2 NA

No. of Ca 2 ions = 2NA

And

No. of Cl  ions = 4NA

Moles of anhydrous CaCl 2 

8.

Ag 2CO3 Initially

0.01 moles

Finally

-

 

2Ag

-

CO2 -

1 O2 2 -

0.02 moles 0.01 moles

0.005 moles

Mass of residue  0.02  108  2.16 g 9.

 

CaCO3

CaO

CO2

Initially

0.1 moles

-

-

Finally

-

0.1 moles = 5.6 g

0.1 moles = 4.4g

VMC/Basic Stoichiometry

15

HWT-Solutions/Chemistry

Vidyamandir Classes 10.

Let the formula of hydrate be BaCl 2 . xH 2O Moles of hydrated BaCl 2 

1.763 208  18x

Moles of anhydrous BaCl 2 = Now,

1.763 1.505  208  18x 208

1.505 208

 x2

Basic Stoichiometry 1.

HWT - 5

3Fe  4H 2O   Fe3O4  4H 2

3 moles 1 mole 4

Initially

3  56g 4  42g 

C  O2

2.

  CO2

1000 83.3 moles = 83.3  32g 12 250   83.3 moles 3 Volume of O2 at STP = 83.3  22.4  1866.6 L Initially

100  9433.35L 21 (As air contains 21 % of O2 by volume) Volume of air at STP = 1866 

3.

2SO2

Initially

10 moles

Finally

-

  2SO3

O2

15 moles 10 moles

10 moles

10 moles of O2 did not enter into combination

4.9 1   0.05 98 20 Moles of = 4  0.05  0.2

6.

Moles of H 3PO4 

7.

Mass of water = 0.04  1  0.04g Moles of water =

0.04  0.0022 18

Molecules of H 2O  0.0022  6  1023  132 1019  1.32  1021

8.

Moles of O2 

3.2  0.1  32

Moles of atoms of O = 0.2

Atoms of O  0.2  6  1023  1.2  1023  0.2N A

8 e s  8  0.2 N A  1.6N A  N A 5

VMC/Basic Stoichiometry

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HWT-Solutions/Chemistry

Vidyamandir Classes 9.

Consider 1 L (1000 Ml) of solution. Moles of NaCl = 3 (As molarity is 3M) Mass of H 2O = Mass of solution – Mass of NaCl  1250  3  58.5 = 1074.5 g

 10.

Molality =

No. of g atms =

3  1000  2.8m 1074.5

2  1023 6  1023

 0.33

Weight = 0.33  32  10.6 g

Basic Stoichiometry 1.

HWT - 6

Meq. of metal = Meq of H 2

1000 

E = 25

0.05 24.62   1000  2 E 24620

2. Initially Finally

x + 5 0.5

2y   9 -

z 4.5

3.

MnO2 

 

4HCl

Initially

2 moles

4 moles

Finally

1 mole

-

 2H 2O 

MnCl 2 -

-

1 mole

% yield of Cl 2 

Cl 2 -

2 moles

1 mole = 22.4L

11.2  100%  50% 22.4

4.

2Fe2S3

6H 2O

3O2

Initially

1 mole

2 moles

3 moles

Finally

1 moles 3

-

2 moles

5.

Let 100g/mol be molar mass of the compound 75.8 1 Moles of atoms of element X  75 24.2  1.5 Moles of atoms of element Y  16 So, formula of compound is XY1.5 or X 2 Y3

6.

Atomic mass 

7.

1.025  1000  1.02  (1000  V)

1.8  1022 1.6  1024

  4Fe(OH)3  6S

4 moles 3

2 moles

 108.36u

V=5

VMC/Basic Stoichiometry

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HWT-Solutions/Chemistry

Vidyamandir Classes 8.

Ca Initially

0.2 moles

Finally

-

2H 2O  

Ca(OH)2

-

H2

-

0.2 moles

Volume of H 2  0.2  22400  4480 cm3

9.

2NO  O2   2NO2 Initially

Finally

Now,

-

x moles 32 -

10 moles 46

x 10 2  x  3.5g 32 46

10. Zn

+

 

I2

ZnI 2

Initially

x moles 65

x moles 254

-

Finally

x   x  65  254  moles  

-

x moles 254

Fraction by weight of original Zn that remains unreacted

x   x  65  254   65     0.74  x   65   65  

Basic Stoichiometry 1.

HWT – 7

21 21  1L  L 100 100 21  0.0093 moles Moles of O2  100  22.4 Volume of O2 

3.

Mg3 N 2  6H 2O   3Mg(OH)2  2NH 3

5.

Initially

1 mole

Finally

-

2 moles

 480   320   1.5     1.2  Total no. of moles  1000 1000     Molarity of final soln.  Total volume (in L) 1 = 0.72 + 0.624 = 1.344 M.

VMC/Basic Stoichiometry

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HWT-Solutions/Chemistry

Vidyamandir Classes 6.

20.8 100 6.1 20.8   0.02 Moles of HCO3 ion = 61 100 Moles of CO2 evolved = 0.02 .

Volume of CO2 evolved = 0.02  24.5 L  0.49 L

Mass of HCO3 ion = 6.1 

(As volume of 1 mole of gas at RTP = 24.5 L) x  0.49  10  4.9  5 . 7.

4Ca 5 (PO4 )3 F  18SiO2  30C   3P4  2CaF2  18CaSiO3  30CO

8.

Initially

0.1 moles

0.36 moles

Finally

0.02 moles

-

0.3 moles

-

-

-

-

0.06 moles

Molar mass = 2  vapour density  2  39.5  79 Let x be valency of element  Formula of chloride is MClx  (3.82  x )  35.5 x  79 

 9.

0.9 moles

x2

Atomic weight of element = 3.82  2  7.64

Let x be valency of element M  Formula of oxide is M 2O x Now, (2  x  14)  (x  16)  44  x  1

Atomic weight of element = 14 1  14

10.

2NH 3

+

5F2

 

Initially

2 moles 17

8 moles 18

Finally

-

8 5  moles 18 17 

+

N 2 F4 -

1 moles 17

3.56  100%  60 % 5.9

Basic Stoichiometry

HWT - 8

A  4B  X

2.

-

1  100g  5.9g 17

% yield 

1.

6HF

Molar mass of X = Molar mass of A + 4 X molar mass of B = 12  4  35.5  154

Ideal gas eqn. PV = nRT We can see that moles SO2  2X moles of O2

molecules of SO2  2X molecules of O2  2N

VMC/Basic Stoichiometry

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HWT-Solutions/Chemistry

Vidyamandir Classes 3 AL  Cl 2   AlCl3 2

4. Initially

0.54 6.67 1  0.02 moles moles  moles  0.05moles 27 133.5 20 Initially three were 0.07 moles of Al and 0.075 moles of Cl 2

Initially three were 0.07 g atoms of Al and 0.15g atoms of Cl 2

Finally

5.

6.

3 O2   AL2O3 2 3 moles Initially 1 mole 4 3   32g  24g 4 2AL 

Acidic meq. used = 25  1  25 Excess of acidic meq. used = 5  1  5

7.

Acidic meq. used for carbonate = Meq. of carbonate = 20

20 

1  1000  E  50 E

Ca(OH)2  H 3PO4   CaHPO4  2H 2O In the above reaction n-factor of H 3PO4 is 2.

98  49 2 Let masses of O2 and N 2 be 1g and 4g

8.

Equivalent weight =

1 NA 32 4 1 Molecules of N 2  N A  N A 28 7 1 1 N A : N A  7 : 32 Ratio of no. of molecules = 32 7 Let x be valency of metal M.  Formula of its chloride = MCl x Molecules of O2 

9.

Moles of MCl x  Now,

M

0.475 0.475   0.01/ x 12x  35.5x 47.5x

x Cl 2   MCl x 2

Initially 0.01/x Finally 0.01.x  0.01x  12x  0.12g 10.

Moles of CO2 evolved =

11  0.25 mole 44

Moles of CO32 ions in the mineral = 0.25 moles

Mass of CO32 ions in the mineral =

% of CO32 in the mineral =

1  60  15g 4

15  100%  50% 30 % of other metal = (100  50  15)%  35%

VMC/Basic Stoichiometry

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HWT-Solutions/Chemistry

Vidyamandir Classes Basic Stoichiometry

1.

HWT - 9

2H 2O2 ( )   2H 2O( )  O2 (g) Initially

1 moles 4 3 1  moles 24 8

Finally

1 100  x 4 1000

 x  2.5

KClO3   KCl 

2.

Initially 1 mole Finally -

3 O2 2 3/2 moles

3 2Al  O2   Al 2O3 2 3 moles Initially 2 Finally 1 mole 3.

4.

5.

Let 100 g/mol be molar mass of the compound. 50 5 Moles of atoms of element X = 10 50  2.5  Moles of atoms of element Y = 20

Formula of the compound is X 5 Y2.5 or X 2 Y

According to the question 10x / 98 9  1000  x = 80.3 910 10  0.5  0.005 Moles of H 2SO4 used  1000

Moles of H  ions used  2  0.005  0.01

Moles of NaOH used = 0.01 (As Na 2SO4 is neutral being a salt of strong acid and strong base)

Mass of NaOH used = 0.01  40  0.4g

6.

For minimum molecular weight, assume that one molecule of the organic compound contains only one S atom. 4  molecular weight  32  Molecular weight = 800. Then 100

7.

Let volumes be 4x L and xL Total no. of m moles (4000x)  (2000x)   1.2M Then final molarity = Total volume (in mL) 5000x

8.

50% Water  Mass of water = Mass of Na 2SO3  126g

9.

126 7  Formula is Na 2SO3 . 7H 2O . 18 Volume of 1L of steam = 0.00006  1000 cm3 = 0.06 cm3

10.

m

Moles of water 

B 1000 0.2 1000     13.88 m 1  B M A 1  0.2 18

VMC/Basic Stoichiometry

21

HWT-Solutions/Chemistry

Vidyamandir Classes Basic Stoichiometry 1.

HWT - 10

(20  1)  (10  0.25)  0.75 40  20

(100  1)  (50  0.25)  0.75 100  50

(40  1)  (20  0.25)  0.75 40  20

(50  1)  (25  0.25)  0.75 50  25

(0.9  100) 90 90  100%   100%   100%  47.36% (0.9  100)  (1  100) 90  100 190

2.

Mass % of H 2SO4 

3.

Let M gm/mol be molar mass of acid 6 12.5  2  1000   10  0.1  1 (Decinormal means normality is 0.1 N) Now M 1000  M = 150

4.

Consider 1000L of sample of hard water

4.75   5.5   1000  200 Total meq. of CaCl 2 and MgCl 2   55.5 47.5   200  50  10g Mass of CaCO3 (meq.  200)  1000  Hardness is 10 ppm 5.

Acidic meq. = (10  0.2)  (30  0.1)  5 

6.

Mass of alcohol in 2nd solution 

7.

Mass = Moles  Molar mass =

8.

Consider 1 L (1000 mL) of soln. 1000 30  1.095   9M Molarity = 36.5 100

10.

Weight of H 2O 

0.61  1000  E

E  122

10  80  0.9  7.2 g 100 90 7.2V  V  0.8  g Mass of alcohol in 1st solution = 100 10 7.2V  V  10mL Now, 7.2  10 1 1022 6  1023

 250  4.14 g

(100  92)  100g  8g 100

VMC/Basic Stoichiometry

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HWT-Solutions/Chemistry