Chem+110+Test+2+Solution+(2012)

School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban CHEM110/CHEM195: General Principles of Chemistry TEST 2: Thursday, 29 March 2012

School of Chemistry and Physics, Ph ysics, University of KwaZulu-Natal, Westville Campus GENERAL PRINCIPLES OF CHEMISTRY - CHEM110/CHEM195 TEST 2 Date: Thursday 29 March 2012

Total marks: 25

Time: 17h45 – 18h30 Examiners: Dr V Paideya Mrs H Govender Compl ete this th is par par t imme imm ediately. IMPORTANT: Comple

Surname:

Initials:

Student No: Tutorial Day: Tutorial Group: Tutor’s Name:

INSTRUCTIONS:

1.

Answer ALL questions.

2.

For Section A which contains the multiple choice questions, qu estions, follow the instructions given in the question.

3.

Calculators may be used but all working must be shown .

4.

Your answers for Section B must be written on the question paper in the spaces  provided. The left-hand pages may be used for extra space or for rough work.

5.

Marks will be deducted for the incorrect use o f significant figures and the omission of  units.

6.

You must write legibly in black or blue ink. Pencils and Tipp-Ex are not allowed .

7.

This test consists of 11 pages including a data sheet and a periodic table.

8.

Please check that you have them all. Question No.

1

2

3

Mark 

1

4

5

Total

School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban CHEM110/CHEM195: General Principles of Chemistry TEST 2: Thursday, 29 March 2012

SECTION A - Multiple Choice Questions 

For each of the following questions, select the correct answer from the list provided.



There is only one correct answer for each question.



Indicate your answer on the multiple choice answe r sheet provided.



Make a dark heavy mark with a HB pencil that fills the block of the appropriate letter  completely.

1.

Hydrogen and nitrogen react to form ammonia according to the reaction, 3H2 + N2 → 2NH3 If 4.0 moles of H2 with 2.0 mol of N2 are reacted, calculate how many moles of NH3 could be produced?

2.

A

0.38 moles

B

2.0 moles

C

2.7 mol

D

4.0 mol

(1)

A 5.72 g sample if magnesium nitride is reacted with excess water in the following reaction: Mg3 N2 + 3H2O → 2NH3 + 3MgO

Determine the theoretical yield of MgO?

A

6.85 g

B

2.28 g

C

0.170 g

D

4.75 g

(1) 2

School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban CHEM110/CHEM195: General Principles of Chemistry TEST 2: Thursday, 29 March 2012

3.

The actual yield of MgO in question (2) above is 4.90 g. What is the percentage yield for  this reaction?

4.

A

97.4%

B

71.5%

C

49.4%

D

87.3%

(1)

Which of the following are combination reactions?

i) CH4(g) + O2(g) → CO2(g) + H2O(l ) ii) CaO(s) + CO 2(g) → CaCO3(s) iii) Mg(s) + O2(g) → MgO(s) iv) PbCO3(s) → PbO(s) + CO2(g)

A

i), ii), and iii)

B

ii) and iii)

C

i), ii), iii), and iv)

D

iv) only

(1)

3

School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban CHEM110/CHEM195: General Principles of Chemistry TEST 2: Thursday, 29 March 2012

5.

In this reaction 3Mg + 2HNO3(dilute) + 6H

+

2+



3Mg

+ 2NO + 4H2O

the magnesium acts as a reducing agent. How many electrons does each magnesium atom lose?

6.

A

1

B

2

C

3

D

4

(1)

In this reaction, which substance behaves as the oxidizing agent? Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O

7.

A

Pb

B

PbSO4

C

PbO2

D

H2SO4

(1)

Equal volumes of all gases, under the same conditions of temperature and pressure, contain ver y nearl y the same number of molecules. This is

A

Avogadro’s Principle.

B

Boyle’s Law.

C

Dalton’s Law.

D

Gay-Lussac’s Law.

(1) 4

School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban CHEM110/CHEM195: General Principles of Chemistry TEST 2: Thursday, 29 March 2012

8.

9.

Which of the following pairs of aqueous solutions will give a precipitate when mixed?

A

LiClO4 and Ba(OH)2

B

AgNO3 and Ca(ClO4)2

C

K 3PO4 and CaCl2

D

NaCl and K(CH3COO)

Consider the following equation and select the correct net ionic equation. -

+

2+

Li + Br  + Pb

10.

(1)

-

+ NO3 →

A

PbBr 2(s) + LiNO3(s)

B

Pb

C

Pb(NO3)2(s) + Li + 2Br 

D

PbBr 2(s) + Li + NO3

2+

-

+ 2Br  + LiNO3(s) -

+

+

-

(1)

In the reaction between aqueous potassium hydroxide and aqueous nitric acid, which are the so-called "spectator" ions?

+

-

A

K  and OH

B

H and NO3

C

K  and NO3

D

H and OH

+

-

+

-

+

-

(1) [10] End of Section A

5

School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban CHEM110/CHEM195: General Principles of Chemistry TEST 2: Thursday, 29 March 2012

SECTION B QUESTION 1

Copper(II) chloride reacts with sodium nitrate to form copper(II) nitrate and sodium chloride according to the equation below: CuCl2 + 2NaNO3 → Cu(NO3)2 + 2NaCl 1.1

If 15.0g of copper(II) chloride reacts with 20.0g of sodium nitrate, calculate how much of  sodium chloride is formed? (2½)

nNaCl = nCuCl2 2

1

Moles of NaCl = 2 x Moles of CuCl2 Mass of NaCl = 2 x nCuCl2 x molar mass of NaCl -1

= 2 x 15.0g/ 134.45g mol

-1

x 58.44g mol

= 13.0g nNaCl = nNaNO3 2

2

Moles of NaCl = Moles of NaNO3 Mass of NaCl = nNaNO3 x molar mass of NaCl -1

-1

= 20.0g/85.0g mol x 58.44g mol = 13.8g

The limiting reagent determines how much product is formed so 13.0g of NaCl is formed.

1.2

Which is the limiting reagent in the question above (1.1)? (½) Copper(II) chloride

6

School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban CHEM110/CHEM195: General Principles of Chemistry TEST 2: Thursday, 29 March 2012

1.3

Determine how much of the excess reagent is left over at the end of the reaction. (1) Mass of NaNO3 = nNaCl x molar mass of NaNO3 -1

-1

= 13g/58.44g mol x 85g mol = 18.9g

Excess reagent remaining = 20g – 18.9g = 0.88g

QUESTION 2

Balance the following equation in basic medium. Show the balanced oxidation and reduction half-reactions and the balanced overall reaction −

2−

−

Cr(OH) 3(s) + ClO3 (aq) → CrO4 (aq) + Cl (aq) (4) Oxidation half-reaction: Cr(OH) 3 + H2O

CrO4



2(Cr(OH)3 + H2O

2−

+

CrO4



−

+ 5H + 3e 2−

+

−

+ 5H + 3e )

Reduction half-reaction: −

+

−

ClO3 + 6H + 6e



−

Cl + 3H2O

Balanced overall reaction: -

2Cr(OH)3(s) + ClO3 (aq) −

−

−

+ 4OH

2Cr(OH)3 + ClO3 + 4OH 2Cr(OH)3 + ClO3

-

−

2−

−

+

2CrO4 (aq) + Cl (aq) + H2O(l) + 4H (aq)







2CrO4 2CrO4

2−

2−

−

+

−

+ Cl + H2O + 4H + 4OH −

+ Cl + H2O + 4H2O

−

2−

−

2Cr(OH)3(s) + ClO3 (aq) + 4OH (aq)  2CrO4 (aq) + Cl (aq) + 5H2O(l)

7

School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban CHEM110/CHEM195: General Principles of Chemistry TEST 2: Thursday, 29 March 2012

QUESTION 3

A 0.3126 g sample of oxalic acid, H2C2O4, requires 26.21 mL of a particular concentration of   NaOH(aq) solution to complete the reaction given below. Calculate the molarity of the NaOH solution. -1

The molar mass of oxalic acid is 90.04 g mol . 2NaOH(aq) + H2C2O4(aq)



Na2C2O4(aq) + 2H2O(aq) (2)

From the balanced equation the reacting ratio of NaOH to C2O4 i.e moles of NaOH / moles of C2O4

2-

2-

is 2:1

= 2/1 2-

i.e MNaOH × VNaOH = 2/1 × moles of C2O4 2-

i.e. MNaOH = 2/1 × moles of C2O4 / VNaOH -1

-3

= 2/1 × (0.3126 g /90.04 g mol ) / 0.02621 dm =

-3

0.2649 mol dm

QUESTION 4

a)

3

o

Calculate the volume, in m , of hydrogen collected over water at 18 C and 725 mmHg when 0.840 g of lithium reacts with water. The balanced equation for the reaction is: 2Li(s) + H2O(l )



H2(g) + 2LiOH(aq) o

The partial pressure of H2O(g) is 15.5 mmHg at 18 C. 3

-1

-1

(3)

Given R = 8.314 m Pa mol K 

 partial pressure of dry hydrogen gas using SI units PH2 = (725 – 15.5) mmHg = 709.5 mmHg/760 mmHg atm = 0.934 atm (convert to Pa) 8

-1

School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban CHEM110/CHEM195: General Principles of Chemistry TEST 2: Thursday, 29 March 2012

PH2

= 0.947 atm x 101325 Pa atm -1 4

= 9.46 x 10 Pa

(ii)

moles of hydrogen gas

nH2 = nLi/2 = 0.840g/ 6.941g mol

-1

x 2 = 0.0605mol

V = nRT/P = (0.0605 x 8.314 m 3 Pa mol-1K -1 x 291.15K )/ 9.46 x 10 4 Pa = 1.55 x 10 -3m3

-1

o

 b) What is the density, in g L , of a sample of oxygen gas at 0.987 atm and 25 C? 3

-1

-1

Given R = 8.314 m Pa mol K 

(2)

Density = MP/RT  5

-1

3

-1

-1

= (1.00 × 10 Pa x 32.00 g mol )/8.314 m Pa mol K  x 298.15 K. -3

= 1291 g m = 1.29 g L

(convert volume from cubic meters to litres)

-1

[15]

End of Section B

9