Secondary 4 Pure Chemistry Stoichiometry Moles Quiz

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Secondary 4 Pure Chemistry Quiz - Stoichiometry Moles

Name: __________________________
Class: __________________________
Date: ___________________________
Score: ________ / 45

Duration: 45 Minutes
Total Marks: 45

Instructions:

1. Answer all questions.
2. Write your answers in the spaces provided.
3. Show all working for calculation questions. Marks may be awarded for correct working even if the final answer is incorrect.
4. Use the following relative atomic masses ($A_r$) unless stated otherwise:
   $H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, S = 32, Cl = 35.5, Ca = 40, Fe = 56, Cu = 63.5, Zn = 65$.
5. Molar volume of gas at room temperature and pressure (r.t.p.) = $24 \text{ dm}^3$.

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Section A: Multiple Choice & Basic Concepts (10 Marks)

1. Which statement about the mole is correct?
A. One mole of any gas occupies $22.4 \text{ dm}^3$ at r.t.p.
B. One mole of any substance contains the same number of particles.
C. The mass of one mole of any substance is always 1 gram.
D. One mole of electrons has a mass of 1 gram.

[1]

2. What is the number of molecules in $0.5 \text{ mol}$ of carbon dioxide ($CO_2$)?
(Let $L$ be the Avogadro constant)
A. $0.5 L$
B. $1.0 L$
C. $1.5 L$
D. $3.0 L$

[1]

3. Which of the following contains the greatest number of atoms?
A. $1 \text{ mol}$ of helium gas ($He$)
B. $1 \text{ mol}$ of oxygen gas ($O_2$)
C. $1 \text{ mol}$ of ammonia gas ($NH_3$)
D. $1 \text{ mol}$ of methane gas ($CH_4$)

[1]

4. What is the empirical formula of a compound with the molecular formula $C_6H_{12}O_6$?
A. $CHO$
B. $CH_2O$
C. $C_2H_4O_2$
D. $C_3H_6O_3$

[1]

5. $20 \text{ cm}^3$ of propane ($C_3H_8$) is burned in excess oxygen. What volume of carbon dioxide is produced? (All volumes measured at the same temperature and pressure)
$C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$
A. $20 \text{ cm}^3$
B. $40 \text{ cm}^3$
C. $60 \text{ cm}^3$
D. $100 \text{ cm}^3$

[1]

6. Calculate the percentage by mass of nitrogen in ammonium nitrate, $NH_4NO_3$.
A. 17.5%
B. 35.0%
C. 46.7%
D. 80.0%

[1]

7. Which solution contains the highest concentration of chloride ions?
A. $1.0 \text{ mol/dm}^3$ $NaCl$
B. $1.0 \text{ mol/dm}^3$ $MgCl_2$
C. $0.5 \text{ mol/dm}^3$ $AlCl_3$
D. $0.5 \text{ mol/dm}^3$ $FeCl_3$

[1]

8. What is the mass of $0.25 \text{ mol}$ of calcium carbonate ($CaCO_3$)?
A. 25 g
B. 50 g
C. 100 g
D. 200 g

[1]

9. In the reaction $2Mg + O_2 \rightarrow 2MgO$, 48 g of magnesium reacts with 32 g of oxygen. Which reactant is in excess?
A. Magnesium
B. Oxygen
C. Neither (stoichiometric amounts)
D. Cannot be determined

[1]

10. A student prepares a solution by dissolving 4.0 g of sodium hydroxide ($NaOH$) in water to make $250 \text{ cm}^3$ of solution. What is the concentration of the solution in $\text{mol/dm}^3$?
A. $0.1 \text{ mol/dm}^3$
B. $0.4 \text{ mol/dm}^3$
C. $1.0 \text{ mol/dm}^3$
D. $16.0 \text{ mol/dm}^3$

[1]

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Section B: Structured Calculations (20 Marks)

11. Iron(III) oxide reacts with carbon monoxide to produce iron and carbon dioxide.
$Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)$

(a) Calculate the relative molecular mass ($M_r$) of iron(III) oxide ($Fe_2O_3$).
[1]
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(b) Calculate the maximum mass of iron that can be produced from 160 g of iron(III) oxide.
[2]
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(c) Calculate the volume of carbon monoxide gas (at r.t.p.) required to react completely with 160 g of iron(III) oxide.
[2]
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12. Hydrated copper(II) sulfate has the formula $CuSO_4 \cdot xH_2O$.
A student heats 5.00 g of the hydrated crystals until all the water of crystallisation is removed. The mass of the remaining anhydrous copper(II) sulfate ($CuSO_4$) is 3.20 g.

(a) Calculate the mass of water lost.
[1]
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(b) Calculate the number of moles of anhydrous $CuSO_4$ remaining. ($M_r$ of $CuSO_4 = 159.5$)
[1]
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(c) Calculate the number of moles of water lost. ($M_r$ of $H_2O = 18$)
[1]
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(d) Determine the value of $x$ in the formula $CuSO_4 \cdot xH_2O$.
[1]
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13. Magnesium reacts with dilute hydrochloric acid according to the equation:
$Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$

In an experiment, 0.12 g of magnesium ribbon is added to $50.0 \text{ cm}^3$ of $0.50 \text{ mol/dm}^3$ hydrochloric acid.

(a) Calculate the number of moles of magnesium used.
[1]
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(b) Calculate the number of moles of $HCl$ present in the solution.
[1]
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(c) Show by calculation which reactant is in excess.
[2]
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(d) Calculate the maximum volume of hydrogen gas produced at r.t.p.
[2]
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14. A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

(a) Calculate the empirical formula of the compound.
[3]
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(b) The relative molecular mass of the compound is 60. Determine the molecular formula.
[1]
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Section C: Application & Analysis (15 Marks)

15. Sodium carbonate reacts with nitric acid to produce sodium nitrate, water, and carbon dioxide.
$Na_2CO_3(s) + 2HNO_3(aq) \rightarrow 2NaNO_3(aq) + H_2O(l) + CO_2(g)$

(a) Explain why the mass of the reaction flask decreases during the reaction if it is not stoppered.
[1]
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(b) In an experiment, excess sodium carbonate is added to $25.0 \text{ cm}^3$ of $2.0 \text{ mol/dm}^3$ nitric acid. Calculate the volume of carbon dioxide produced at r.t.p.
[3]
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(c) If the actual volume of gas collected was $0.50 \text{ dm}^3$, calculate the percentage yield of the reaction.
[2]
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16. Zinc blende is an ore containing zinc sulfide ($ZnS$). It is roasted in air to produce zinc oxide and sulfur dioxide.
$2ZnS(s) + 3O_2(g) \rightarrow 2ZnO(s) + 2SO_2(g)$

(a) Calculate the percentage purity of a sample of zinc blende if 10.0 g of the ore produces 8.1 g of zinc oxide ($ZnO$) upon complete roasting.
($A_r: Zn=65, S=32, O=16$)
[4]
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(b) Suggest one environmental problem caused by the release of sulfur dioxide gas and how it can be prevented.
[2]
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17. A student wants to prepare $250 \text{ cm}^3$ of $0.10 \text{ mol/dm}^3$ sodium chloride solution from solid sodium chloride.

(a) Calculate the mass of sodium chloride required.
[2]
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(b) Describe the steps the student should take to prepare this solution accurately using a volumetric flask.
[3]
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18. Limiting Reagent Analysis.
Aluminium reacts with chlorine gas to form aluminium chloride.
$2Al(s) + 3Cl_2(g) \rightarrow 2AlCl_3(s)$

5.4 g of aluminium is reacted with 14.2 g of chlorine gas.

(a) Calculate the moles of each reactant.
[2]
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(b) Identify the limiting reagent.
[1]
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(c) Calculate the mass of aluminium chloride produced.
[2]
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19. Titration Calculation.
$25.0 \text{ cm}^3$ of potassium hydroxide solution ($KOH$) is neutralised by $20.0 \text{ cm}^3$ of $0.50 \text{ mol/dm}^3$ sulfuric acid ($H_2SO_4$).
$2KOH(aq) + H_2SO_4(aq) \rightarrow K_2SO_4(aq) + 2H_2O(l)$

(a) Calculate the moles of sulfuric acid used.
[1]
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(b) Calculate the moles of potassium hydroxide that reacted.
[1]
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(c) Calculate the concentration of the potassium hydroxide solution in $\text{mol/dm}^3$.
[2]
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20. Gas Stoichiometry.
Ethene ($C_2H_4$) burns in oxygen to form carbon dioxide and water.
$C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)$

$10 \text{ cm}^3$ of ethene is mixed with $50 \text{ cm}^3$ of oxygen and ignited. The mixture is then cooled to room temperature.

(a) Which gas is in excess?
[1]
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(b) Calculate the volume of the excess gas remaining.
[2]
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(c) Calculate the total volume of gas remaining in the mixture after cooling. (Ignore the volume of liquid water).
[2]
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Secondary 4 Pure Chemistry Quiz - Stoichiometry Moles (Answer Key)

1. B
Explanation: One mole of any substance contains the Avogadro constant ($6.02 \times 10^{23}$) of particles. A is incorrect because molar volume is $24 \text{ dm}^3$ at r.t.p. (or $22.4$ at s.t.p., but the question specifies r.t.p. context usually implies 24 in SG syllabus, regardless, B is the definition). C and D are factually incorrect.

2. A
Explanation: Number of molecules = moles $\times L$. $0.5 \times L = 0.5L$.

3. D
Explanation:
A: $1 \times 1 = 1$ mol atoms.
B: $1 \times 2 = 2$ mol atoms.
C: $1 \times 4 = 4$ mol atoms.
D: $1 \times 5 = 5$ mol atoms.
Methane has the most atoms.

4. B
Explanation: Ratio C:H:O is 6:12:6. Divide by highest common factor (6) $\rightarrow$ 1:2:1. Formula $CH_2O$.

5. C
Explanation: Ratio $C_3H_8 : CO_2$ is 1:3. Volume $CO_2 = 3 \times 20 \text{ cm}^3 = 60 \text{ cm}^3$.

6. B
Explanation: $M_r(NH_4NO_3) = 14 + 4(1) + 14 + 3(16) = 80$.
Mass of N = $14 + 14 = 28$.
$\% N = (28/80) \times 100 = 35.0\%$.

7. B
Explanation:
A: $[Cl^-] = 1.0 \text{ M}$.
B: $[Cl^-] = 1.0 \times 2 = 2.0 \text{ M}$.
C: $[Cl^-] = 0.5 \times 3 = 1.5 \text{ M}$.
D: $[Cl^-] = 0.5 \times 3 = 1.5 \text{ M}$.
$MgCl_2$ has the highest concentration.

8. A
Explanation: $M_r(CaCO_3) = 40 + 12 + 3(16) = 100$.
Mass = $0.25 \text{ mol} \times 100 \text{ g/mol} = 25 \text{ g}$.

9. B
Explanation:
Moles Mg = $48/24 = 2$ mol.
Moles $O_2$ = $32/32 = 1$ mol.
Ratio Mg:$O_2$ is 2:1. We have exactly 2 mol Mg and 1 mol $O_2$. They are in stoichiometric proportions.
Correction/Refinement: Wait, let's re-read carefully. "Which reactant is in excess?" If they are stoichiometric, neither is in excess. However, usually, these questions have a trick. Let's re-calculate.
$2Mg + O_2 \rightarrow 2MgO$.
2 mol Mg requires 1 mol $O_2$.
We have 2 mol Mg and 1 mol $O_2$.
Answer C is "Neither".
Self-Correction for Key: The options provided in Q9 were A, B, C, D. C is "Neither". So Answer is C.

10. B
Explanation:
Moles $NaOH = 4.0 / 40 = 0.1$ mol.
Volume = $250 \text{ cm}^3 = 0.25 \text{ dm}^3$.
Concentration = $0.1 / 0.25 = 0.4 \text{ mol/dm}^3$.

11.
(a) $M_r = 2(56) + 3(16) = 112 + 48 = 160$. [1]
(b) Moles $Fe_2O_3 = 160 \text{ g} / 160 \text{ g/mol} = 1.0$ mol.
From equation, 1 mol $Fe_2O_3$ produces 2 mol Fe.
Moles Fe = 2.0 mol.
Mass Fe = $2.0 \times 56 = 112$ g. [2]
(c) From equation, 1 mol $Fe_2O_3$ reacts with 3 mol CO.
Moles CO = 3.0 mol.
Volume CO = $3.0 \times 24 = 72 \text{ dm}^3$. [2]

12.
(a) Mass water = $5.00 - 3.20 = 1.80$ g. [1]
(b) Moles $CuSO_4 = 3.20 / 159.5 \approx 0.020$ mol. [1]
(c) Moles $H_2O = 1.80 / 18 = 0.10$ mol. [1]
(d) Ratio $H_2O : CuSO_4 = 0.10 : 0.020 = 5 : 1$. So $x = 5$. [1]

13.
(a) Moles Mg = $0.12 / 24 = 0.005$ mol. [1]
(b) Moles HCl = $0.50 \times (50/1000) = 0.025$ mol. [1]
(c) Ratio Mg:HCl is 1:2.
0.005 mol Mg requires $0.005 \times 2 = 0.010$ mol HCl.
We have 0.025 mol HCl, which is greater than 0.010 mol.
Therefore, HCl is in excess. [2]
(d) Limiting reagent is Mg.
Moles $H_2$ produced = Moles Mg = 0.005 mol.
Volume $H_2 = 0.005 \times 24 = 0.12 \text{ dm}^3$ (or $120 \text{ cm}^3$). [2]

14.
(a)
C: $40.0/12 = 3.33$
H: $6.7/1 = 6.7$
O: $53.3/16 = 3.33$
Divide by smallest (3.33):
C: 1, H: 2, O: 1.
Empirical Formula: $CH_2O$. [3]
(b) $M_r(CH_2O) = 12+2+16 = 30$.
Ratio = $60/30 = 2$.
Molecular Formula: $C_2H_4O_2$. [1]

15.
(a) Carbon dioxide gas escapes from the flask. [1]
(b) Moles $HNO_3 = 2.0 \times (25/1000) = 0.05$ mol.
Ratio $Na_2CO_3 : HNO_3$ is 1:2.
Moles $CO_2$ produced = $1/2 \times$ moles $HNO_3 = 0.025$ mol.
Volume $CO_2 = 0.025 \times 24 = 0.60 \text{ dm}^3$. [3]
(c) % Yield = $(\text{Actual} / \text{Theoretical}) \times 100$.
$(0.50 / 0.60) \times 100 = 83.3\%$. [2]

16.
(a) Moles $ZnO = 8.1 / (65+16) = 8.1 / 81 = 0.1$ mol.
From equation, 2 mol $ZnS$ produces 2 mol $ZnO$ (1:1 ratio).
Moles $ZnS$ reacted = 0.1 mol.
Mass pure $ZnS = 0.1 \times (65+32) = 0.1 \times 97 = 9.7$ g.
% Purity = $(9.7 / 10.0) \times 100 = 97\%$. [4]
(b) Problem: Acid rain / Respiratory problems.
Prevention: Flue gas desulfurization / React with calcium carbonate/lime. [2]

17.
(a) Moles needed = $0.10 \times 0.250 = 0.025$ mol.
Mass = $0.025 \times (23+35.5) = 0.025 \times 58.5 = 1.4625$ g (accept 1.46 g). [2]
(b) 1. Weigh 1.46 g of NaCl.
2. Dissolve in a beaker with some distilled water.
3. Transfer to $250 \text{ cm}^3$ volumetric flask (rinse beaker).
4. Add distilled water to the mark. [3]

18.
(a) Moles Al = $5.4 / 27 = 0.2$ mol.
Moles $Cl_2 = 14.2 / (35.5 \times 2) = 14.2 / 71 = 0.2$ mol. [2]
(b) Ratio Al:$Cl_2$ is 2:3.
0.2 mol Al requires $0.2 \times (3/2) = 0.3$ mol $Cl_2$.
We only have 0.2 mol $Cl_2$.
So $Cl_2$ is limiting. [1]
(c) Moles $AlCl_3$ produced based on $Cl_2$.
Ratio $Cl_2 : AlCl_3$ is 3:2.
Moles $AlCl_3 = 0.2 \times (2/3) = 0.1333$ mol.
$M_r AlCl_3 = 27 + 3(35.5) = 133.5$.
Mass = $0.1333 \times 133.5 = 17.8$ g. [2]

19.
(a) Moles $H_2SO_4 = 0.50 \times (20/1000) = 0.01$ mol. [1]
(b) Ratio KOH:$H_2SO_4$ is 2:1.
Moles KOH = $2 \times 0.01 = 0.02$ mol. [1]
(c) Conc KOH = Moles / Volume($dm^3$).
Volume = $25 \text{ cm}^3 = 0.025 \text{ dm}^3$.
Conc = $0.02 / 0.025 = 0.8 \text{ mol/dm}^3$. [2]

20.
(a) Ratio $C_2H_4 : O_2$ is 1:3.
$10 \text{ cm}^3$ ethene requires $30 \text{ cm}^3$ oxygen.
We have $50 \text{ cm}^3$ oxygen.
Oxygen is in excess. [1]
(b) Excess $O_2 = 50 - 30 = 20 \text{ cm}^3$. [2]
(c) $CO_2$ produced: Ratio $C_2H_4 : CO_2$ is 1:2.
Vol $CO_2 = 2 \times 10 = 20 \text{ cm}^3$.
Total gas = Excess $O_2$ + $CO_2$ (water is liquid).
Total = $20 + 20 = 40 \text{ cm}^3$. [2]