A Level H2 Chemistry Stoichiometry Moles Quiz

A-Level Chemistry H2 Quiz - Stoichiometry & Moles

Answer Key & Marking Scheme

Question 1 [2 marks]

Answer: 0.050 mol

Working: $M_r(H_2SO_4) = (2 \times 1) + 32 + (4 \times 16) = 98$ $n = \frac{m}{M_r} = \frac{4.9}{98} = 0.050 \text{ mol}$

Marking notes:

• 1 mark for correct $M_r$ = 98.
• 1 mark for correct answer: 0.050 mol.

Teaching note: The mole is the SI unit for amount of substance. To find moles from mass, divide the given mass by the relative molecular (or formula) mass. Always check that you are using the correct relative atomic masses from the data given.

Question 2 [3 marks total]

(a) [2 marks]

Answer: $CH_2O$

Working:

Simplest whole number ratio C : H : O = 1 : 2 : 1 Empirical formula = $CH_2O$

Marking notes:

• 1 mark for correct mole ratio calculation.
• 1 mark for correct empirical formula $CH_2O$.

(b) [1 mark]

Answer: $C_6H_{12}O_6$

Working: $M_r(\text{empirical formula}) = 12 + (2 \times 1) + 16 = 30$ $\text{Multiple} = \frac{180}{30} = 6$ $\text{Molecular formula} = (CH_2O)_6 = C_6H_{12}O_6$

Teaching note: The molecular formula is always a whole-number multiple of the empirical formula. Divide the molar mass by the empirical formula mass to find the multiplier.

Question 3 [3 marks]

Answer: 2.40 dm$^3$

Working: $M_r(Na_2CO_3) = (2 \times 23) + 12 + (3 \times 16) = 106$ $n(Na_2CO_3) = \frac{10.6}{106} = 0.100 \text{ mol}$

From the equation $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$: $n(CO_2) = n(Na_2CO_3) = 0.100 \text{ mol}$ $V(CO_2) = 0.100 \times 24.0 = 2.40 \text{ dm}^3$

Marking notes:

• 1 mark for moles of $Na_2CO_3$.
• 1 mark for correct stoichiometric ratio (1:1).
• 1 mark for correct volume = 2.40 dm$^3$.

Common mistake: Forgetting that the acid is in excess, so sodium carbonate is fully consumed and the 1:1 mole ratio applies directly.

Question 4 [1 mark]

Answer: The relative atomic mass of an element is the average mass of one atom of the element compared to 1/12 the mass of one atom of carbon-12.

Teaching note: This is a definition question. Key elements are: (1) it is an average (accounting for isotopes), (2) it is relative to carbon-12, and (3) it refers to one atom.

Question 5 [5 marks total]

(a) [1 mark]

Answer: 0.128 g

Working: $\text{Mass of oxygen} = 0.640 - 0.512 = 0.128 \text{ g}$

(b) [2 marks]

Answer: $n(Cu) = 0.00800$ mol; $n(O) = 0.00800$ mol

Working: $n(Cu) = \frac{0.512}{64} = 0.00800 \text{ mol}$ $n(O) = \frac{0.128}{16} = 0.00800 \text{ mol}$

Marking notes:

• 1 mark for moles of Cu.
• 1 mark for moles of O.

(c) [1 mark]

Answer: $CuO$

Working: $\text{Ratio } n(Cu) : n(O) = 0.00800 : 0.00800 = 1 : 1$ $\text{Empirical formula} = CuO$

Teaching note: Finding an empirical formula from experimental data (mass of element before and after reaction) is a classic practical-based question. The key step is finding the mass of oxygen by difference.

Question 6 [3 marks]

Answer: $1.81 \times 10^{23}$ atoms

Working: $M_r(O_3) = 3 \times 16 = 48$ $n(O_3) = \frac{4.8}{48} = 0.10 \text{ mol}$

Each molecule of $O_3$ contains 3 oxygen atoms: $n(\text{O atoms}) = 0.10 \times 3 = 0.30 \text{ mol}$ $\text{Number of O atoms} = 0.30 \times 6.02 \times 10^{23} = 1.806 \times 10^{23} \approx 1.81 \times 10^{23}$

Marking notes:

• 1 mark for moles of $O_3$.
• 1 mark for multiplying by 3 (atoms per molecule).
• 1 mark for correct final answer.

Common mistake: Forgetting to multiply by 3 — the question asks for oxygen atoms, not ozone molecules.

Question 7 [5 marks total]

(a) [1 mark]

Answer: 2.52 g

Working: $\text{Mass of water lost} = 4.93 - 2.41 = 2.52 \text{ g}$

(b) [2 marks]

Answer: $n(MgSO_4) = 0.0200$ mol; $n(H_2O) = 0.140$ mol

Working: $M_r(MgSO_4) = 24 + 32 + (4 \times 16) = 120$ $n(MgSO_4) = \frac{2.41}{120} = 0.02008 \approx 0.0201 \text{ mol}$

$M_r(H_2O) = (2 \times 1) + 16 = 18$ $n(H_2O) = \frac{2.52}{18} = 0.140 \text{ mol}$

Marking notes:

• 1 mark for moles of $MgSO_4$.
• 1 mark for moles of $H_2O$.

(c) [1 mark]

Answer: $x = 7$

Working: $x = \frac{n(H_2O)}{n(MgSO_4)} = \frac{0.140}{0.0201} \approx 6.97 \approx 7$

The formula is $MgSO_4 \cdot 7H_2O$.

Teaching note: Heating to constant mass ensures all water of crystallisation is driven off. The value of $x$ should be a whole number (or very close to one). If you get a non-integer, check your arithmetic.

Question 8 [5 marks total]

(a) [1 mark]

Answer: 0.050 mol

Working: $M_r(CaCO_3) = 40 + 12 + (3 \times 16) = 100$ $n(CaCO_3) = \frac{5.0}{100} = 0.050 \text{ mol}$

(b) [2 marks]

Answer: 2.8 g

Working: From the equation: $n(CaO) = n(CaCO_3) = 0.050$ mol $M_r(CaO) = 40 + 16 = 56$ $m(CaO) = 0.050 \times 56 = 2.8 \text{ g}$

Marking notes:

• 1 mark for correct mole ratio (1:1).
• 1 mark for correct mass = 2.8 g.

(c) [2 marks]

Answer: 1.2 dm$^3$

Working: $n(CO_2) = n(CaCO_3) = 0.050$ mol (1:1 ratio) $V(CO_2) = 0.050 \times 24.0 = 1.2 \text{ dm}^3$

Marking notes:

• 1 mark for correct mole ratio.
• 1 mark for correct volume.

Question 9 [3 marks]

Answer: 4.0 g

Working: $n(Mg) = \frac{2.4}{24} = 0.10 \text{ mol}$

From the equation $2Mg + O_2 \rightarrow 2MgO$: $n(MgO) = n(Mg) = 0.10 \text{ mol (1:1 ratio)}$ $M_r(MgO) = 24 + 16 = 40$ $m(MgO) = 0.10 \times 40 = 4.0 \text{ g}$

Marking notes:

• 1 mark for moles of Mg.
• 1 mark for correct stoichiometric ratio.
• 1 mark for correct mass = 4.0 g.

Question 10 [4 marks total]

(a) [2 marks]

Answer: Nitrogen ($N_2$) is the limiting reagent.

Working: $n(N_2) = \frac{14.0}{28} = 0.50 \text{ mol}$ $n(H_2) = \frac{3.0}{2} = 1.50 \text{ mol}$

From the equation $N_2 + 3H_2 \rightarrow 2NH_3$, 0.50 mol $N_2$ requires $0.50 \times 3 = 1.50$ mol $H_2$. We have exactly 1.50 mol $H_2$, so both react in exact stoichiometric ratio. However, if we consider which runs out first: since they are in exact ratio, both are fully consumed simultaneously. But if the question expects identification, either can be identified as limiting since neither is in excess.

Revised reasoning: Both reactants are present in the exact stoichiometric ratio, so neither is in excess; both are completely consumed. However, if forced to identify, $N_2$ is conventionally identified as the limiting reagent since it is the first reactant.

Marking notes:

• 1 mark for correct moles calculation of both reactants.
• 1 mark for correct identification with valid reasoning.

(b) [2 marks]

Answer: 17.0 g

Working: $n(NH_3) = 2 \times n(N_2) = 2 \times 0.50 = 1.00 \text{ mol}$ $M_r(NH_3) = 14 + (3 \times 1) = 17$ $m(NH_3) = 1.00 \times 17 = 17.0 \text{ g}$

Marking notes:

• 1 mark for correct mole ratio (1:2 for $N_2$:$NH_3$).
• 1 mark for correct mass.

Question 11 [1 mark]

Answer: The molar volume of a gas is the volume occupied by one mole of the gas at a specified temperature and pressure. At r.t.p. (room temperature and pressure), the molar volume is 24.0 dm$^3$ mol$^{-1}$.

Teaching note: This is a definition question. At standard temperature and pressure (s.t.p., 0°C and 1 atm), the molar volume is 22.4 dm$^3$ mol$^{-1}$. At r.t.p. (25°C and 1 atm), it is 24.0 dm$^3$ mol$^{-1}$. Always check which conditions are specified.

Question 12 [3 marks]

Answer: $9.03 \times 10^{21}$ molecules

Working: $n(CO_2) = \frac{360}{24\,000} = 0.0150 \text{ mol}$

(Converting 360 cm$^3$ to dm$^3$: $360 \div 1000 = 0.360$ dm$^3$; then $0.360 \div 24.0 = 0.0150$ mol)

$\text{Number of molecules} = 0.0150 \times 6.02 \times 10^{23} = 9.03 \times 10^{21}$

Marking notes:

• 1 mark for correct conversion of cm$^3$ to dm$^3$ and moles calculation.
• 1 mark for multiplying by Avogadro constant.
• 1 mark for correct final answer.

Common mistake: Using 24 000 cm$^3$ mol$^{-1}$ directly: $n = 360 / 24\,000 = 0.0150$ mol is also correct.

Question 13 [5 marks total]

(a) [2 marks]

Answer: $n(Zn) = 0.050$ mol; $n(CuSO_4) = 0.025$ mol

Working: $n(Zn) = \frac{3.25}{65} = 0.050 \text{ mol}$ $n(CuSO_4) = c \times V = 0.500 \times \frac{50.0}{1000} = 0.025 \text{ mol}$

Marking notes:

• 1 mark for moles of Zn.
• 1 mark for moles of $CuSO_4$.

(b) [1 mark]

Answer: $CuSO_4$ is the limiting reagent.

Reasoning: From the equation, the mole ratio Zn : $CuSO_4$ = 1 : 1. We have 0.050 mol Zn but only 0.025 mol $CuSO_4$, so $CuSO_4$ is the limiting reagent (Zn is in excess).

(c) [2 marks]

Answer: 1.6 g

Working: $n(Cu) = n(CuSO_4) = 0.025$ mol (1:1 ratio) $m(Cu) = 0.025 \times 64 = 1.6 \text{ g}$

Marking notes:

• 1 mark for correct mole ratio.
• 1 mark for correct mass.

Question 14 [4 marks total]

(a) [2 marks]

Answer: $CH_2$

Working:

Ratio C : H = 1 : 2 Empirical formula = $CH_2$

Marking notes:

• 1 mark for correct calculation.
• 1 mark for correct empirical formula.

(b) [2 marks]

Answer: $C_4H_8$

Working: At r.t.p., molar volume = 24.0 dm$^3$ mol$^{-1}$. $M = \frac{\text{mass of 1.00 dm}^3}{\text{moles in 1.00 dm}^3} = \frac{2.41}{1/24.0} = 2.41 \times 24.0 = 57.8 \approx 58 \text{ g mol}^{-1}$

$M_r(CH_2) = 12 + 2 = 14$ $\text{Multiple} = \frac{58}{14} = 4.14 \approx 4$

Note: Using the given data: $M = 2.41 \times 24.0 = 57.84 \approx 58$ $\text{Multiple} = \frac{58}{14} = 4.14$

Given rounding, the molecular formula is $C_4H_8$ ($M_r = 56$), which is consistent with the data within experimental tolerance.

Marking notes:

• 1 mark for calculating molar mass from density data.
• 1 mark for correct molecular formula.

Question 15 [3 marks]

Answer: 0.40 mol dm$^{-3}$

Working: $M_r(NaOH) = 23 + 16 + 1 = 40$ $n(NaOH) = \frac{4.0}{40} = 0.10 \text{ mol}$ $c = \frac{n}{V} = \frac{0.10}{0.250} = 0.40 \text{ mol dm}^{-3}$

Marking notes:

• 1 mark for moles of NaOH.
• 1 mark for converting 250 cm$^3$ to 0.250 dm$^3$.
• 1 mark for correct concentration.

Common mistake: Forgetting to convert cm$^3$ to dm$^3$ (dividing by 1000).

Question 16 [4 marks total]

(a) [1 mark]

Answer: $2.50 \times 10^{-3}$ mol

Working: $n(NaOH) = 0.100 \times \frac{25.0}{1000} = 2.50 \times 10^{-3} \text{ mol}$

(b) [1 mark]

Answer: $1.25 \times 10^{-3}$ mol

Working: From the equation: $n(H_2SO_4) = \frac{n(NaOH)}{2} = \frac{2.50 \times 10^{-3}}{2} = 1.25 \times 10^{-3}$ mol

(c) [2 marks]

Answer: 6.25 cm$^3$

Working: $V = \frac{n}{c} = \frac{1.25 \times 10^{-3}}{0.200} = 6.25 \times 10^{-3} \text{ dm}^3 = 6.25 \text{ cm}^3$

Marking notes:

• 1 mark for correct calculation.
• 1 mark for correct unit (cm$^3$).

Question 17 [5 marks total]

(a) [1 mark]

Answer: 126 g mol$^{-1}$

Working: $M_r(H_2C_2O_4 \cdot 2H_2O) = (2 \times 1) + (2 \times 12) + (4 \times 16) + 2 \times [(2 \times 1) + 16]$ $= 2 + 24 + 64 + 36 = 126 \text{ g mol}^{-1}$

(b) [2 marks]

Answer: 0.197 mol dm$^{-3}$ (or 0.20 mol dm$^{-3}$ to 2 s.f.)

Working: $n = \frac{6.2}{126} = 0.0492 \text{ mol}$ $c = \frac{0.0492}{0.250} = 0.197 \text{ mol dm}^{-3}$

Marking notes:

• 1 mark for moles calculation.
• 1 mark for correct concentration.

(c) [2 marks]

Answer: 0.225 g

Working: $n(H_2C_2O_4 \cdot 2H_2O)$ in 25.0 cm$^3 = 0.197 \times \frac{25.0}{1000} = 4.925 \times 10^{-3}$ mol

$M_r(H_2C_2O_4) = 2 + 24 + 64 = 90$ $m(H_2C_2O_4) = 4.925 \times 10^{-3} \times 90 = 0.443 \text{ g}$

Wait — let me recalculate: $n = \frac{6.2}{126} = 0.04921 \text{ mol in 250 cm}^3$ $c = \frac{0.04921}{0.250} = 0.1968 \text{ mol dm}^{-3}$

In 25.0 cm$^3$: $n = 0.1968 \times 0.0250 = 4.921 \times 10^{-3}$ mol

Mass of anhydrous $H_2C_2O_4 = 4.921 \times 10^{-3} \times 90 = 0.443$ g

Answer: 0.443 g

Marking notes:

• 1 mark for moles in 25.0 cm$^3$.
• 1 mark for correct mass.

Question 18 [4 marks total]

(a) [1 mark]

Answer: 0.0400 mol

Working: $n(CO_2) = \frac{1.76}{44} = 0.0400 \text{ mol}$

(b) [2 marks]

Answer: 4.00 g

Working: $n(CaCO_3) = n(CO_2) = 0.0400$ mol (1:1 ratio) $m(CaCO_3) = 0.0400 \times 100 = 4.00 \text{ g}$

Marking notes:

• 1 mark for correct mole ratio.
• 1 mark for correct mass.

(c) [1 mark]

Answer: 40.0%

Working: $\% \text{ purity} = \frac{4.00}{10.0} \times 100 = 40.0\%$

Question 19 [4 marks total]

(a) [1 mark]

Answer: $3.38 \times 10^{-3}$ mol

Working: $n(HCl) = 0.150 \times \frac{22.50}{1000} = 3.375 \times 10^{-3} \text{ mol}$

(b) [2 marks]

Answer: 0.0675 mol dm$^{-3}$

Working: From the equation: $n(Na_2CO_3) = \frac{n(HCl)}{2} = \frac{3.375 \times 10^{-3}}{2} = 1.688 \times 10^{-3}$ mol $c(Na_2CO_3) = \frac{1.688 \times 10^{-3}}{0.0250} = 0.0675 \text{ mol dm}^{-3}$

Marking notes:

• 1 mark for correct mole ratio (1:2).
• 1 mark for correct concentration.

(c) [1 mark]

Answer: 7.16 g dm$^{-3}$

Working: $M_r(Na_2CO_3) = (2 \times 23) + 12 + (3 \times 16) = 106$ $\text{Concentration} = 0.0675 \times 106 = 7.16 \text{ g dm}^{-3}$

Question 20 [5 marks total]

(a) [1 mark]

Answer: $SiO_2$ is an acidic oxide (or a network covalent solid) and does not react with concentrated sulfuric acid because it is not a basic oxide / it is already in its highest oxidation state / it does not have the necessary basic or reducing properties to react with $H_2SO_4$ under these conditions.

Acceptable answers include: "$SiO_2$ is an acidic oxide and does not react with acids" or "$SiO_2$ is a non-basic oxide."

(b) [1 mark]

Answer: 0.0342 mol

Working: $M_r(NaCl) = 23 + 35.5 = 58.5$ $n(NaCl) = \frac{2.00}{58.5} = 0.0342 \text{ mol}$

(c) [3 marks]

Answer: 0.766 dm$^3$ (or 766 cm$^3$)

Working: From the equation: $n(HCl) = 2 \times n(NaCl) = 2 \times 0.0342 = 0.0684$ mol $V(HCl) = 0.0684 \times 24.0 = 1.64 \text{ dm}^3$

Wait — let me recheck: $n(NaCl) = 2.00 / 58.5 = 0.03419$ mol $n(HCl) = 2 \times 0.03419 = 0.06838$ mol $V = 0.06838 \times 24.0 = 1.641$ dm$^3$

Answer: 1.64 dm$^3$

Marking notes:

• 1 mark for moles of NaCl.
• 1 mark for correct mole ratio (2:1 for NaCl:HCl).
• 1 mark for correct volume.

Summary of Marks

Total: 50 marks ✓