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

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School of Chemistry and Physics, 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

IMPORTANT: Complete this part immediately.

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, 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 of 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 4 5 Total

Mark




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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 answer 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?

A 0.38 moles

B 2.0 moles

C 2.7 mol

D 4.0 mol (1)

2. A 5.72 g sample if magnesium nitride is reacted with excess water in the following

reaction:

Mg3N2 + 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)




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

this reaction?

A 97.4%

B 71.5%

C 49.4%

D 87.3% (1)

4. Which of the following are combination reactions?

i) CH4(g) + O2(g) → CO2(g) + H2O(l)

ii) CaO(s) + CO2(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)




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5. In this reaction

3Mg + 2HNO3(dilute) + 6H+ 3Mg

2+ + 2NO + 4H2O

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

lose?

A 1

B 2

C 3

D 4 (1)

6. In this reaction, which substance behaves as the oxidizing agent?

Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O

A Pb

B PbSO4

C PbO2

D H2SO4 (1)

7. Equal volumes of all gases, under the same conditions of temperature and pressure,

contain very nearly 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)




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8. 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 K3PO4 and CaCl2

D NaCl and K(CH3COO) (1)

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

Li+

+ Br- + Pb

2+ + NO3

- →

A PbBr2(s) + LiNO3(s)

B Pb2+

+ 2Br- + LiNO3(s)

C Pb(NO3)2(s) + Li+ + 2Br

-

D PbBr2(s) + Li+

+ NO3

- (1)

10. 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




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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

= 2 x 15.0g/ 134.45g mol-1

x 58.44g mol-1

= 13.0g

nNaCl = nNaNO3

2 2

Moles of NaCl = Moles of NaNO3

Mass of NaCl = nNaNO3 x molar mass of NaCl

= 20.0g/85.0g mol-1

x 58.44g mol-1

= 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




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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

= 13g/58.44g mol-1

x 85g mol-1

= 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

Cr(OH)3(s) + ClO3−(aq) → CrO4

2−(aq) + Cl

−(aq)

(4)

Oxidation half-reaction:

Cr(OH)3 + H2O CrO42−

+ 5H+ + 3e

−

2(Cr(OH)3 + H2O CrO42−

+ 5H+ + 3e

− )

Reduction half-reaction:

ClO3− + 6H

+ + 6e

− Cl

− + 3H2O

Balanced overall reaction:

2Cr(OH)3(s) + ClO3-(aq) 2CrO4

2−(aq) + Cl

−(aq) + H2O(l) + 4H

+(aq)

2Cr(OH)3 + ClO3− + 4OH

− 2CrO4

2− + Cl

− + H2O + 4H

+ + 4OH

−

2Cr(OH)3 + ClO3− + 4OH

- 2CrO4

2− + Cl

− + H2O + 4H2O

2Cr(OH)3(s) + ClO3−(aq) + 4OH

−(aq) 2CrO4

2−(aq) + Cl

−(aq) + 5H2O(l)




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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.

The molar mass of oxalic acid is 90.04 g mol-1

.

2NaOH(aq) + H2C2O4(aq) Na2C2O4(aq) + 2H2O(aq)

(2)

From the balanced equation the reacting ratio of NaOH to C2O42-

is 2:1

i.e moles of NaOH / moles of C2O42-

= 2/1

i.e MNaOH × VNaOH = 2/1 × moles of C2O42-

i.e. MNaOH = 2/1 × moles of C2O42-

/ VNaOH

= 2/1 × (0.3126 g /90.04 g mol-1

) / 0.02621 dm-3

= 0.2649 mol dm-3

QUESTION 4

a) Calculate the volume, in m3, of hydrogen collected over water at 18

oC 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)

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

Given R = 8.314 m3 Pa mol

-1 K

-1 (3)

partial pressure of dry hydrogen gas using SI units

PH2 = (725 – 15.5) mmHg

= 709.5 mmHg/760 mmHg atm-1

= 0.934 atm (convert to Pa)




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PH2 = 0.947 atm x 101325 Pa atm-1

= 9.46 x 104 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 m3 Pa mol

-1K

-1 x 291.15K )/ 9.46 x 10

4 Pa

= 1.55 x 10-3

m3

b) What is the density, in g L-1

, of a sample of oxygen gas at 0.987 atm and 25 oC?

Given R = 8.314 m3 Pa mol

-1 K

-1 (2)

Density = MP/RT

= (1.00 × 105 Pa x 32.00 g mol

-1)/8.314 m

3 Pa mol

-1 K

-1 x 298.15 K.

= 1291 g m-3

(convert volume from cubic meters to litres)

= 1.29 g L

-1

[15]

End of Section B

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

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