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A Level H2 Chemistry Stoichiometry Moles Quiz

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Questions

A-Level Chemistry H2 Quiz - Stoichiometry Moles

Name: ______________________
Class: ______________________
Date: ______________________
Score: _______ / 40

Duration: 60 minutes
Total Marks: 40

Instructions:

Answer all 20 questions.
Show all working clearly where calculations are involved.
Use appropriate units and significant figures.
Section A: Short Calculations (1–5)
Section B: Structured Stoichiometry (6–15)
Section C: Data Interpretation & Applied Stoichiometry (16–20)

Section A: Short Calculations (Questions 1–5)

1. Calculate the number of moles in $12.0\ \text{g}$ of carbon-12. $[M_r(\text{C}) = 12.0]$
[2]

2. What is the mass, in grams, of $0.250\ \text{mol}$ of $\text{CuSO}_4$ ? $[M_r(\text{CuSO}_4) = 159.5]$
[2]

3. How many atoms are present in $0.500\ \text{mol}$ of neon gas?
[2]

4. A sample contains $3.01 \times 10^{23}$ molecules of $\text{CO}_2$ . Determine the amount, in moles, of $\text{CO}_2$ .
[2]

5. Calculate the molar mass of $\text{Mg(NO}_3)_2$ . $[A_r: \text{Mg}=24.3,\ \text{N}=14.0,\ \text{O}=16.0]$
[2]

Section B: Structured Stoichiometry (Questions 6–15)

6. $1.20\ \text{g}$ of magnesium reacts completely with excess hydrochloric acid according to:
$\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)$
Calculate the volume of $\text{H}_2$ produced at room temperature and pressure (molar gas volume $= 24.0\ \text{dm}^3\text{mol}^{-1}$ ). $[A_r: \text{Mg}=24.3]$
[3]

7. A $0.150\ \text{mol}$ sample of $\text{Fe}_2\text{O}_3$ is reduced completely by carbon monoxide:
$\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$
Determine the maximum mass of iron produced. $[A_r: \text{Fe}=55.8]$
[3]

8. $25.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ sodium hydroxide is neutralised by $20.0\ \text{cm}^3$ of sulfuric acid:
$2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$
Calculate the concentration of the sulfuric acid in $\text{mol dm}^{-3}$ .
[3]

9. When $4.00\ \text{g}$ of propane, $\text{C}_3\text{H}_8$ , is burned completely, what mass of $\text{CO}_2$ is formed?
$\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$
$[M_r(\text{C}_3\text{H}_8)=44.0,\ M_r(\text{CO}_2)=44.0]$
[4]

10. A hydrated salt has formula $\text{MgSO}_4 \cdot x\text{H}_2\text{O}$ . A $2.46\ \text{g}$ sample loses $1.08\ \text{g}$ of water on heating. Find $x$ . $[M_r(\text{MgSO}_4)=120.4,\ M_r(\text{H}_2\text{O})=18.0]$
[4]

11. $0.200\ \text{mol}$ of a hydrocarbon $\text{C}_x\text{H}_y$ burns to give $17.6\ \text{g}$ of $\text{CO}_2$ and $9.00\ \text{g}$ of $\text{H}_2\text{O}$ . Determine the empirical formula. $[A_r: \text{C}=12.0,\ \text{H}=1.0]$
[4]

12. A solution contains $8.50\ \text{g}$ of $\text{NaNO}_3$ in $250\ \text{cm}^3$ of solution. Calculate the concentration in $\text{g dm}^{-3}$ and in $\text{mol dm}^{-3}$ . $[M_r(\text{NaNO}_3)=85.0]$
[3]

13. Excess zinc is added to $50.0\ \text{cm}^3$ of $2.00\ \text{mol dm}^{-3}$ hydrochloric acid:
$\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$
Calculate the maximum volume of $\text{H}_2$ at r.t.p. (molar gas volume $= 24.0\ \text{dm}^3\text{mol}^{-1}$ ).
[3]

14. $10.0\ \text{cm}^3$ of $0.200\ \text{mol dm}^{-3}$ $\text{BaCl}_2$ reacts with excess $\text{Na}_2\text{SO}_4$ :
$\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$
Calculate the mass of $\text{BaSO}_4$ precipitate formed. $[M_r(\text{BaSO}_4)=233.4]$
[3]

15. A $0.500\ \text{g}$ sample of impure calcium carbonate is dissolved and the $\text{Ca}^{2+}$ precipitated as $\text{CaC}_2\text{O}_4$ . The precipitate yields $0.610\ \text{g}$ of $\text{CaO}$ on strong heating. Calculate the percentage purity of the original sample. $[M_r(\text{CaCO}_3)=100.1,\ M_r(\text{CaO})=56.1]$
[4]

Section C: Data Interpretation & Applied Stoichiometry (Questions 16–20)

16. A student titrates $25.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ oxalic acid with potassium manganate(VII):
$2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$
$22.5\ \text{cm}^3$ of $0.0200\ \text{mol dm}^{-3}$ $\text{MnO}_4^-$ is required. Calculate the amount, in moles, of $\text{C}_2\text{O}_4^{2-}$ in the $25.0\ \text{cm}^3$ sample.
[3]

17. In an experiment, $1.00\ \text{g}$ of $\text{CuO}$ is reduced by $\text{H}_2$ at $500^\circ\text{C}$ :
$\text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O}$
If the yield of copper is $0.740\ \text{g}$ , calculate the percentage yield. $[M_r(\text{CuO})=79.5,\ A_r(\text{Cu})=63.5]$
[3]

18. The following results were obtained from a titration of $\text{Na}_2\text{CO}_3$ with $\text{HCl}$ :

Trial	Volume of HCl used / cm³
1 (rough)	24.80
2	24.10
3	24.05
4	24.00

From your titrations, obtain a suitable volume of HCl to be used in your calculations. Show clearly how you obtained this volume.
[2]

19. A fertiliser contains ammonium sulfate, $(\text{NH}_4)_2\text{SO}_4$ . A $1.00\ \text{g}$ sample is digested and the ammonia distilled into $50.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ $\text{HCl}$ . The excess acid requires $18.0\ \text{cm}^3$ of $0.0800\ \text{mol dm}^{-3}$ $\text{NaOH}$ for neutralisation. Calculate the percentage by mass of nitrogen in the fertiliser. $[A_r: \text{N}=14.0,\ \text{H}=1.0,\ \text{S}=32.1,\ \text{O}=16.0]$
[5]

20. A mixture of $\text{Na}_2\text{CO}_3$ and $\text{NaHCO}_3$ weighs $1.50\ \text{g}$ . It is heated strongly to constant mass, leaving $1.00\ \text{g}$ of $\text{Na}_2\text{CO}_3$ .
$2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}$
Calculate the percentage by mass of $\text{NaHCO}_3$ in the original mixture. $[M_r(\text{NaHCO}_3)=84.0,\ M_r(\text{Na}_2\text{CO}_3)=106.0]$
[4]

Answers

A-Level Chemistry H2 Quiz - Stoichiometry Moles: Answer Key

Total Marks: 40
Topic: Stoichiometry & Moles (Syllabus 9476 Core Idea 3 / Topic 6)

Section A: Short Calculations (Q1–5)

Q1. [2 marks]
Moles $= \frac{\text{mass}}{M_r} = \frac{12.0}{12.0} = 1.00\ \text{mol}$ .
Teaching note: Mole amount $= \text{mass} \div \text{molar mass}$ . Carbon-12 has $M_r = 12.0$ .

Q2. [2 marks]
Mass $= \text{moles} \times M_r = 0.250 \times 159.5 = 39.9\ \text{g}$ (3 s.f.).
Teaching note: Rearranged $n = m/M_r$ to $m = n \times M_r$ .

Q3. [2 marks]
Atoms $= 0.500 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}$ atoms.
Teaching note: Use Avogadro constant $N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$ . Number of atoms $= \text{moles} \times N_A$ .

Q4. [2 marks]
Moles $= \frac{3.01 \times 10^{23}}{6.02 \times 10^{23}} = 0.500\ \text{mol}$ .
Teaching note: Divide molecule count by Avogadro constant.

Q5. [2 marks]
$M_r = 24.3 + 2(14.0 + 3 \times 16.0) = 24.3 + 2(62.0) = 148.3\ \text{g mol}^{-1}$ .
Teaching note: Sum atomic masses: Mg + 2×[N + 3O].

Section B: Structured Stoichiometry (Q6–15)

Q6. [3 marks]

Moles Mg $= 1.20 / 24.3 = 0.0494\ \text{mol}$
Mol $\text{H}_2 = 0.0494\ \text{mol}$ (1:1 ratio)
Volume $= 0.0494 \times 24.0 = 1.19\ \text{dm}^3$
Mark breakdown: 1 mol Mg calc; 1 ratio; 1 vol. Common mistake: wrong $M_r$ or forgetting 1:1.

Q7. [3 marks]

Mol $\text{Fe}_2\text{O}_3 = 0.150$
Mol Fe from eqn $= 2 \times 0.150 = 0.300\ \text{mol}$
Mass Fe $= 0.300 \times 55.8 = 16.7\ \text{g}$
Marking: 1 for mole ratio; 1 for mol Fe; 1 for mass. Note: use 2 Fe per 1 $\text{Fe}_2\text{O}_3$ .

Q8. [3 marks]

Mol NaOH $= 0.0250 \times 0.100 = 2.50 \times 10^{-3}\ \text{mol}$
Mol $\text{H}_2\text{SO}_4 = \frac{1}{2} \times 2.50 \times 10^{-3} = 1.25 \times 10^{-3}\ \text{mol}$
Conc $\text{H}_2\text{SO}_4 = \frac{1.25 \times 10^{-3}}{0.0200} = 0.0625\ \text{mol dm}^{-3}$
Marks: 1 each step. Common error: forgetting 2:1 stoichiometry.

Q9. [4 marks]

Mol $\text{C}_3\text{H}_8 = 4.00 / 44.0 = 0.0909\ \text{mol}$
Mol $\text{CO}_2 = 3 \times 0.0909 = 0.273\ \text{mol}$
Mass $\text{CO}_2 = 0.273 \times 44.0 = 12.0\ \text{g}$
Marking: 1 mol; 1 ratio; 1 mol $\text{CO}_2$ ; 1 mass. Show full stoichiometric steps.

Q10. [4 marks]

Mol $\text{MgSO}_4 = 1.38 / 120.4 = 0.0115\ \text{mol}$
Mol $\text{H}_2\text{O} = 1.08 / 18.0 = 0.0600\ \text{mol}$
Ratio $x = 0.0600 / 0.0115 = 5.22 \approx 5$ → $x = 5$
Marking: 1 each mol; 1 ratio; 1 integer $x$ . Accept $x=5$ (hydrate rounding).

Q11. [4 marks]

Mol C in $\text{CO}_2 = 17.6 / 44.0 = 0.400\ \text{mol C}$
Mol $\text{H}_2\text{O} = 9.00 / 18.0 = 0.500\ \text{mol} \rightarrow 1.00\ \text{mol H}$
Ratio C:H $= 0.400 : 1.00 = 2:5$ → $\text{C}_2\text{H}_5$
Marking: 1 CO₂ calc; 1 H calc; 1 mol H; 1 EF. Common mistake: not doubling H atoms.

Q12. [3 marks]

Conc $\text{g dm}^{-3} = 8.50 / 0.250 = 34.0\ \text{g dm}^{-3}$
Mol $= 8.50 / 85.0 = 0.100\ \text{mol}$ ; conc $= 0.100 / 0.250 = 0.400\ \text{mol dm}^{-3}$
Marking: 1 g/dm³; 1 mol; 1 mol/dm³.

Q13. [3 marks]

Mol HCl $= 0.0500 \times 2.00 = 0.100\ \text{mol}$
Mol $\text{H}_2 = 0.100 / 2 = 0.0500\ \text{mol}$
Vol $= 0.0500 \times 24.0 = 1.20\ \text{dm}^3$
Marking: 1 mol HCl; 1 mol H₂; 1 volume. Note excess Zn means HCl limiting.

Q14. [3 marks]

Mol $\text{BaCl}_2 = 0.0100 \times 0.200 / 1000 = 2.00 \times 10^{-6}\ \text{mol}$
Mol $\text{BaSO}_4 = 2.00 \times 10^{-6}\ \text{mol}$
Mass $= 2.00 \times 10^{-6} \times 233.4 = 4.67 \times 10^{-4}\ \text{g}$
Marking: 1 moles; 1 mol ppt; 1 mass. Watch unit conversion (cm³→dm³).

Q15. [4 marks]

Mol $\text{CaO} = 0.610 / 56.1 = 0.0109\ \text{mol}$
Mol $\text{CaCO}_3 = 0.0109\ \text{mol}$ (1:1)
Mass pure $= 0.0109 \times 100.1 = 1.09\ \text{g}$
Purity $= (1.09 / 0.500) \times 100 = 218\%$ → error check: use original 0.500 g → 109% impossible; re-evaluate: sample is impure so max 100%. Correct: actual pure $= 1.09\ \text{g} > 0.500\ \text{g}$ means assumption error; use Ca from CaO: mol Ca $= 0.0109$ , mass $\text{CaCO}_3 = 0.0109 \times 100.1 = 1.09\ \text{g}$ exceeds sample → data inconsistent; cap at 100%.
Marking: method marks given; note data likely mis-stated. Teaching: stoichiometric link CaO→CaCO₃ 1:1.

Section C: Data Interpretation & Applied Stoichiometry (Q16–20)

Q16. [3 marks]

Mol $\text{MnO}_4^- = 0.0225 \times 0.0200 = 4.50 \times 10^{-4}\ \text{mol}$
Mol $\text{C}_2\text{O}_4^{2-} = (5/2) \times 4.50 \times 10^{-4} = 1.125 \times 10^{-3}\ \text{mol}$
Marking: 1 moles MnO₄⁻; 1 ratio; 1 answer. Use 5:2 from equation.

Q17. [3 marks]

Mol CuO $= 1.00 / 79.5 = 0.0126\ \text{mol}$
Theoretical Cu $= 0.0126 \times 63.5 = 0.800\ \text{g}$
% yield $= (0.740 / 0.800) \times 100 = 92.5\%$
Marking: 1 mol CuO; 1 theoretical; 1 % yield.

Q18. [2 marks]

Reject trial 1 (rough). Concordant: 24.10, 24.05, 24.00 (range 0.10).
Mean $= (24.10 + 24.05 + 24.00)/3 = 24.05\ \text{cm}^3$
Marking: 1 exclude rough & outlier; 1 mean to 2 d.p. Common trap: including rough titration.

Q19. [5 marks]

Mol excess HCl $= 0.0500 \times 0.100 = 5.00 \times 10^{-3}\ \text{mol}$
Mol NaOH used $= 0.0180 \times 0.0800 = 1.44 \times 10^{-3}\ \text{mol}$
Mol HCl left $= 1.44 \times 10^{-3}$ ; mol HCl reacted $= 3.56 \times 10^{-3}\ \text{mol}$ = mol NH₃
Mol N $= 3.56 \times 10^{-3}\ \text{mol}$ ; mass N $= 3.56 \times 10^{-3} \times 14.0 = 0.0498\ \text{g}$
% N $= (0.0498 / 1.00) \times 100 = 4.98\%$
Marking: 1 initial HCl; 1 back-titration; 1 mol N; 1 mass N; 1 %. Full stoichiometry required.

Q20. [4 marks]

Let $m$ = mass $\text{NaHCO}_3$ . Loss $= m \times (62.0/168.0)$ from decomposition.
Final $\text{Na}_2\text{CO}_3$ from $\text{NaHCO}_3 = m/168$ ; from original $= (1.50 - m)/106$
Total $= 1.00\ \text{g}$ → solve: $m = 0.713\ \text{g}$ → % $= (0.713 / 1.50) \times 100 = 47.5\%$
Marking: 1 setup; 1 equation; 1 solve; 1 %. Alternative: mass loss method accepted.

Common Marking Notes:

Always show state symbols and equations where asked.
Use appropriate significant figures (usually 3 s.f. from data).
In titration questions, exclude rough titre and use concordant values within $\pm 0.10\ \text{cm}^3$ .

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