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A Level H2 Chemistry Stoichiometry Moles Quiz

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Questions

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A-Level Chemistry H2 Quiz - Stoichiometry Moles

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 45 minutes Total Marks: 50

Instructions:

  • Answer ALL questions in the spaces provided.
  • Show all working clearly; marks are awarded for method.
  • Use appropriate significant figures in your final answers.
  • State units where applicable.
  • A calculator may be used.
  • The Data Booklet is relevant to some questions.

Section A: Basic Mole Concepts (Questions 1–5)

[10 marks]

1. Calculate the number of moles of sodium hydroxide in 250 cm³ of 0.400 mol dm⁻³ NaOH solution.

[2 marks]

2. A sample of magnesium contains 3.01 × 10²³ atoms. Calculate the amount, in moles, of magnesium present.

[2 marks]

3. Calculate the mass of potassium nitrate, KNO₃, required to prepare 500 cm³ of a 0.200 mol dm⁻³ solution.

[2 marks]

4. A compound has the empirical formula CH₂O and a relative molecular mass of 180. Determine its molecular formula.

[2 marks]

5. Calculate the volume occupied by 0.500 mol of nitrogen gas at room temperature and pressure (r.t.p.), where the molar volume is 24.0 dm³ mol⁻¹.

[2 marks]

Section B: Stoichiometric Calculations (Questions 6–10)

[12 marks]

6. Calcium carbonate decomposes on heating according to the equation:

CaCO₃(s) → CaO(s) + CO₂(g)

Calculate the mass of calcium oxide produced when 25.0 g of calcium carbonate is heated strongly.

[3 marks]

7. Magnesium reacts with hydrochloric acid:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Calculate the volume of hydrogen gas produced at r.t.p. when 1.20 g of magnesium reacts completely with excess hydrochloric acid.

[3 marks]

8. 2.00 g of an unknown metal carbonate, M₂CO₃, was dissolved in water and made up to 250 cm³. 25.0 cm³ of this solution required 20.0 cm³ of 0.100 mol dm⁻³ HCl for complete neutralisation.

M₂CO₃ + 2HCl → 2MCl + CO₂ + H₂O

Calculate the relative atomic mass of M and identify the metal.

[3 marks]

9. A student prepared a standard solution by dissolving 5.30 g of anhydrous sodium carbonate, Na₂CO₃, in water and making up to 1.00 dm³. Calculate the concentration of this solution in mol dm⁻³.

[3 marks]

10. 10.0 cm³ of a solution of sulfuric acid, H₂SO₄, was diluted to 250 cm³. 25.0 cm³ of the diluted acid required 22.5 cm³ of 0.100 mol dm⁻³ NaOH for complete neutralisation.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Calculate the concentration of the original sulfuric acid solution.

[3 marks]

Section C: Advanced Applications (Questions 11–15)

[14 marks]

11. A hydrocarbon contains 85.7% carbon by mass. Its relative molecular mass is 56.0.

(a) Determine the empirical formula of the hydrocarbon. [2 marks]

(b) Determine its molecular formula. [1 mark]

(c) Draw and name two possible structural isomers of this hydrocarbon. [2 marks]

12. In an experiment, 3.27 g of zinc was added to 50.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid.

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

(a) Determine which reactant is in excess. Show your working. [3 marks]

(b) Calculate the mass of zinc chloride formed. [2 marks]

13. A sample of hydrated magnesium sulfate, MgSO₄·xH₂O, has a mass of 6.15 g. After heating to constant mass, 3.00 g of anhydrous MgSO₄ remains.

Calculate the value of x in MgSO₄·xH₂O.

[3 marks]

14. 0.500 g of an impure sample of sodium chloride was dissolved in water. Excess silver nitrate solution was added to precipitate all the chloride ions as silver chloride. The precipitate was filtered, dried, and weighed. The mass of AgCl obtained was 1.10 g.

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Calculate the percentage purity of the sodium chloride sample.

[3 marks]

15. A gaseous compound contains 30.4% nitrogen and 69.6% oxygen by mass. 0.250 g of the gas occupies 82.0 cm³ at 100 °C and 101 kPa.

(a) Determine the empirical formula of the compound. [2 marks]

(b) Calculate the relative molecular mass of the compound using the ideal gas equation. [2 marks]

(c) Hence, determine the molecular formula. [1 mark]

Section D: Integrated Problem Solving (Questions 16–20)

[14 marks]

16. A student carried out a titration to determine the concentration of a solution of ethanedioic acid, H₂C₂O₄. 25.0 cm³ of the acid required 23.80 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII) for complete oxidation.

2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

Calculate the concentration of the ethanedioic acid solution in g dm⁻³.

[4 marks]

17. A mixture contains sodium chloride and sodium bromide. 0.500 g of the mixture was dissolved in water and treated with excess silver nitrate solution. The total mass of silver halide precipitate formed was 0.950 g.

Calculate the percentage by mass of sodium bromide in the mixture.

[4 marks]

18. When 2.00 g of a Group 2 metal, M, was added to excess water, 1.12 dm³ of hydrogen gas was collected at r.t.p.

M(s) + 2H₂O(l) → M(OH)₂(aq) + H₂(g)

Identify metal M by calculating its relative atomic mass.

[3 marks]

19. A compound of phosphorus and chlorine contains 14.9% phosphorus by mass. 0.200 g of the compound was vaporised and found to occupy 35.5 cm³ at 150 °C and 100 kPa.

Determine the molecular formula of the compound.

[3 marks]

20. A 1.00 g sample of an alloy containing only aluminium and zinc was dissolved in excess hydrochloric acid. The hydrogen gas produced was collected over water at 25 °C and 100 kPa. The volume of gas collected was 1.20 dm³. The saturated vapour pressure of water at 25 °C is 3.17 kPa.

2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g) Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

Calculate the percentage by mass of aluminium in the alloy.

[4 marks]

END OF QUIZ

Answers

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

Total Marks: 50

Section A: Basic Mole Concepts (Questions 1–5)

1. Calculate the number of moles of sodium hydroxide in 250 cm³ of 0.400 mol dm⁻³ NaOH solution. [2 marks]

Answer: n = c × V V = 250 cm³ = 0.250 dm³ n = 0.400 × 0.250 = 0.100 mol ✓

Marking:

  • 1 mark: correct conversion of cm³ to dm³
  • 1 mark: correct calculation and answer with units

2. A sample of magnesium contains 3.01 × 10²³ atoms. Calculate the amount, in moles, of magnesium present. [2 marks]

Answer: n = number of particles ÷ Avogadro constant n = 3.01 × 10²³ ÷ 6.02 × 10²³ = 0.500 mol ✓

Marking:

  • 1 mark: correct use of Avogadro constant
  • 1 mark: correct answer with units

3. Calculate the mass of potassium nitrate, KNO₃, required to prepare 500 cm³ of a 0.200 mol dm⁻³ solution. [2 marks]

Answer: n = c × V = 0.200 × 0.500 = 0.100 mol Mᵣ(KNO₃) = 39.1 + 14.0 + (3 × 16.0) = 101.1 Mass = n × Mᵣ = 0.100 × 101.1 = 10.1 g ✓

Marking:

  • 1 mark: correct moles calculation
  • 1 mark: correct mass with units

4. A compound has the empirical formula CH₂O and a relative molecular mass of 180. Determine its molecular formula. [2 marks]

Answer: Mᵣ of empirical formula CH₂O = 12.0 + (2 × 1.0) + 16.0 = 30.0 n = 180 ÷ 30.0 = 6 Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ ✓

Marking:

  • 1 mark: correct calculation of empirical formula mass and factor
  • 1 mark: correct molecular formula

5. Calculate the volume occupied by 0.500 mol of nitrogen gas at room temperature and pressure (r.t.p.), where the molar volume is 24.0 dm³ mol⁻¹. [2 marks]

Answer: Volume = n × molar volume Volume = 0.500 × 24.0 = 12.0 dm³ ✓

Marking:

  • 1 mark: correct formula
  • 1 mark: correct answer with units

Section B: Stoichiometric Calculations (Questions 6–10)

6. CaCO₃(s) → CaO(s) + CO₂(g). Calculate the mass of calcium oxide produced when 25.0 g of calcium carbonate is heated strongly. [3 marks]

Answer: Mᵣ(CaCO₃) = 40.1 + 12.0 + (3 × 16.0) = 100.1 n(CaCO₃) = 25.0 ÷ 100.1 = 0.2498 mol Mole ratio CaCO₃ : CaO = 1 : 1 n(CaO) = 0.2498 mol Mᵣ(CaO) = 40.1 + 16.0 = 56.1 Mass of CaO = 0.2498 × 56.1 = 14.0 g ✓

Marking:

  • 1 mark: correct moles of CaCO₃
  • 1 mark: correct mole ratio application
  • 1 mark: correct final mass with units (accept 14.0–14.1 g)

7. Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). Calculate the volume of hydrogen gas produced at r.t.p. when 1.20 g of magnesium reacts completely with excess hydrochloric acid. [3 marks]

Answer: Aᵣ(Mg) = 24.3 n(Mg) = 1.20 ÷ 24.3 = 0.04938 mol Mole ratio Mg : H₂ = 1 : 1 n(H₂) = 0.04938 mol Volume of H₂ = 0.04938 × 24.0 = 1.19 dm³ ✓

Marking:

  • 1 mark: correct moles of Mg
  • 1 mark: correct mole ratio
  • 1 mark: correct volume with units

8. M₂CO₃ + 2HCl → 2MCl + CO₂ + H₂O. 2.00 g of M₂CO₃ in 250 cm³; 25.0 cm³ required 20.0 cm³ of 0.100 mol dm⁻³ HCl. Calculate Aᵣ(M) and identify the metal. [3 marks]

Answer: n(HCl) in titre = 0.100 × 0.0200 = 0.00200 mol Mole ratio M₂CO₃ : HCl = 1 : 2 n(M₂CO₃) in 25.0 cm³ = 0.00200 ÷ 2 = 0.00100 mol n(M₂CO₃) in 250 cm³ = 0.00100 × 10 = 0.0100 mol Mᵣ(M₂CO₃) = mass ÷ moles = 2.00 ÷ 0.0100 = 200 Mᵣ(M₂CO₃) = 2Aᵣ(M) + 12.0 + (3 × 16.0) = 2Aᵣ(M) + 60.0 2Aᵣ(M) = 200 − 60.0 = 140 Aᵣ(M) = 70.0; Metal is gallium, Ga ✓

Marking:

  • 1 mark: correct moles of HCl and M₂CO₃ in titre
  • 1 mark: correct scaling to total moles and Mᵣ
  • 1 mark: correct Aᵣ and identification (accept Ga or gallium)

9. 5.30 g of anhydrous Na₂CO₃ dissolved in water and made up to 1.00 dm³. Calculate the concentration in mol dm⁻³. [3 marks]

Answer: Mᵣ(Na₂CO₃) = (2 × 23.0) + 12.0 + (3 × 16.0) = 106.0 n(Na₂CO₃) = 5.30 ÷ 106.0 = 0.0500 mol Concentration = n ÷ V = 0.0500 ÷ 1.00 = 0.0500 mol dm⁻³ ✓

Marking:

  • 1 mark: correct Mᵣ
  • 1 mark: correct moles
  • 1 mark: correct concentration with units

10. 10.0 cm³ H₂SO₄ diluted to 250 cm³; 25.0 cm³ of diluted acid required 22.5 cm³ of 0.100 mol dm⁻³ NaOH. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate concentration of original H₂SO₄. [3 marks]

Answer: n(NaOH) = 0.100 × 0.0225 = 0.00225 mol Mole ratio H₂SO₄ : NaOH = 1 : 2 n(H₂SO₄) in 25.0 cm³ = 0.00225 ÷ 2 = 0.001125 mol Concentration of diluted H₂SO₄ = 0.001125 ÷ 0.0250 = 0.0450 mol dm⁻³ Dilution factor = 250 ÷ 10.0 = 25 Concentration of original H₂SO₄ = 0.0450 × 25 = 1.13 mol dm⁻³ ✓

Marking:

  • 1 mark: correct moles of NaOH and H₂SO₄ in titre
  • 1 mark: correct concentration of diluted acid
  • 1 mark: correct original concentration with units

Section C: Advanced Applications (Questions 11–15)

11. Hydrocarbon: 85.7% C, Mᵣ = 56.0. [5 marks]

(a) Empirical formula: [2 marks] Mass of H = 100 − 85.7 = 14.3%

Element%÷ AᵣRatio
C85.785.7 ÷ 12.0 = 7.147.14 ÷ 7.14 = 1
H14.314.3 ÷ 1.0 = 14.314.3 ÷ 7.14 = 2
Empirical formula = CH₂ ✓

(b) Molecular formula: [1 mark] Mᵣ(CH₂) = 14.0; n = 56.0 ÷ 14.0 = 4 Molecular formula = C₄H₈ ✓

(c) Two structural isomers: [2 marks]

  • But-1-ene: CH₂=CHCH₂CH₃
  • But-2-ene: CH₃CH=CHCH₃ (Accept any two correct isomers of C₄H₈ with correct names and structures.)

Marking:

  • (a): 1 mark for correct mole ratio working, 1 mark for CH₂
  • (b): 1 mark for C₄H₈
  • (c): 1 mark each for correct structure and name (max 2)

12. 3.27 g Zn + 50.0 cm³ of 2.00 mol dm⁻³ HCl. Zn + 2HCl → ZnCl₂ + H₂. [5 marks]

(a) Limiting/excess reagent: [3 marks] n(Zn) = 3.27 ÷ 65.4 = 0.0500 mol n(HCl) = 2.00 × 0.0500 = 0.100 mol Mole ratio Zn : HCl = 1 : 2 HCl required for 0.0500 mol Zn = 0.0500 × 2 = 0.100 mol HCl available = 0.100 mol Neither is in excess; they are in exactly stoichiometric amounts. ✓ (Alternative: Zn required for 0.100 mol HCl = 0.0500 mol; Zn available = 0.0500 mol. Same conclusion.)

(b) Mass of ZnCl₂ formed: [2 marks] n(ZnCl₂) = n(Zn) = 0.0500 mol (1:1 ratio) Mᵣ(ZnCl₂) = 65.4 + (2 × 35.5) = 136.4 Mass = 0.0500 × 136.4 = 6.82 g ✓

Marking:

  • (a): 1 mark for moles of Zn, 1 mark for moles of HCl, 1 mark for correct conclusion with reasoning
  • (b): 1 mark for correct moles of ZnCl₂, 1 mark for correct mass

13. MgSO₄·xH₂O: 6.15 g → 3.00 g anhydrous MgSO₄. Find x. [3 marks]

Answer: Mass of water lost = 6.15 − 3.00 = 3.15 g Mᵣ(MgSO₄) = 24.3 + 32.1 + (4 × 16.0) = 120.4 n(MgSO₄) = 3.00 ÷ 120.4 = 0.02492 mol n(H₂O) = 3.15 ÷ 18.0 = 0.175 mol Ratio H₂O : MgSO₄ = 0.175 ÷ 0.02492 = 7.02 x = 7 ✓

Marking:

  • 1 mark: correct mass of water
  • 1 mark: correct moles of MgSO₄ and H₂O
  • 1 mark: correct value of x (accept 7)

14. 0.500 g impure NaCl → 1.10 g AgCl. Ag⁺ + Cl⁻ → AgCl. Calculate % purity. [3 marks]

Answer: Mᵣ(AgCl) = 107.9 + 35.5 = 143.4 n(AgCl) = 1.10 ÷ 143.4 = 0.007671 mol n(Cl⁻) = n(AgCl) = 0.007671 mol (1:1) n(NaCl) = n(Cl⁻) = 0.007671 mol Mᵣ(NaCl) = 23.0 + 35.5 = 58.5 Mass of pure NaCl = 0.007671 × 58.5 = 0.4488 g % purity = (0.4488 ÷ 0.500) × 100 = 89.8% ✓

Marking:

  • 1 mark: correct moles of AgCl
  • 1 mark: correct mass of pure NaCl
  • 1 mark: correct percentage

15. Gas: 30.4% N, 69.6% O; 0.250 g occupies 82.0 cm³ at 100 °C, 101 kPa. [5 marks]

(a) Empirical formula: [2 marks]

Element%÷ AᵣRatio
N30.430.4 ÷ 14.0 = 2.172.17 ÷ 2.17 = 1
O69.669.6 ÷ 16.0 = 4.354.35 ÷ 2.17 = 2
Empirical formula = NO₂ ✓

(b) Relative molecular mass using ideal gas equation: [2 marks] pV = nRT p = 101 kPa = 101000 Pa; V = 82.0 cm³ = 8.20 × 10⁻⁵ m³; T = 100 + 273 = 373 K n = pV ÷ RT = (101000 × 8.20 × 10⁻⁵) ÷ (8.31 × 373) = 0.00267 mol Mᵣ = mass ÷ n = 0.250 ÷ 0.00267 = 93.6 ✓

(c) Molecular formula: [1 mark] Mᵣ(NO₂) = 14.0 + (2 × 16.0) = 46.0 n = 93.6 ÷ 46.0 = 2.03 ≈ 2 Molecular formula = N₂O₄ ✓

Marking:

  • (a): 1 mark for working, 1 mark for NO₂
  • (b): 1 mark for correct n, 1 mark for Mᵣ (accept 92–94)
  • (c): 1 mark for N₂O₄

Section D: Integrated Problem Solving (Questions 16–20)

16. 25.0 cm³ H₂C₂O₄ required 23.80 cm³ of 0.0200 mol dm⁻³ KMnO₄. 2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. Calculate [H₂C₂O₄] in g dm⁻³. [4 marks]

Answer: n(MnO₄⁻) = 0.0200 × 0.02380 = 4.76 × 10⁻⁴ mol Mole ratio MnO₄⁻ : H₂C₂O₄ = 2 : 5 n(H₂C₂O₄) = (5 ÷ 2) × 4.76 × 10⁻⁴ = 1.19 × 10⁻³ mol Concentration = n ÷ V = 1.19 × 10⁻³ ÷ 0.0250 = 0.0476 mol dm⁻³ Mᵣ(H₂C₂O₄) = (2 × 1.0) + (2 × 12.0) + (4 × 16.0) = 90.0 Concentration in g dm⁻³ = 0.0476 × 90.0 = 4.28 g dm⁻³ ✓

Marking:

  • 1 mark: correct moles of MnO₄⁻
  • 1 mark: correct moles of H₂C₂O₄ using mole ratio
  • 1 mark: correct concentration in mol dm⁻³
  • 1 mark: correct conversion to g dm⁻³

17. 0.500 g mixture (NaCl + NaBr) → 0.950 g AgX precipitate. Calculate % NaBr. [4 marks]

Answer: Let mass of NaBr = x g; mass of NaCl = (0.500 − x) g Mᵣ(NaCl) = 58.5; Mᵣ(NaBr) = 102.9 Mᵣ(AgCl) = 143.4; Mᵣ(AgBr) = 187.8

n(NaCl) = (0.500 − x) ÷ 58.5; n(NaBr) = x ÷ 102.9 Mass of AgCl = n(NaCl) × 143.4 = 143.4(0.500 − x) ÷ 58.5 Mass of AgBr = n(NaBr) × 187.8 = 187.8x ÷ 102.9

Total mass = [143.4(0.500 − x) ÷ 58.5] + [187.8x ÷ 102.9] = 0.950

143.4(0.500 − x) ÷ 58.5 = 2.451(0.500 − x) = 1.226 − 2.451x 187.8x ÷ 102.9 = 1.825x

1.226 − 2.451x + 1.825x = 0.950 1.226 − 0.626x = 0.950 −0.626x = −0.276 x = 0.441 g

% NaBr = (0.441 ÷ 0.500) × 100 = 88.2% ✓

Marking:

  • 1 mark: correct algebraic setup with masses and Mᵣ values
  • 1 mark: correct equation linking masses
  • 1 mark: correct solving for x
  • 1 mark: correct percentage

18. 2.00 g Group 2 metal M + excess H₂O → 1.12 dm³ H₂ at r.t.p. M + 2H₂O → M(OH)₂ + H₂. Identify M. [3 marks]

Answer: n(H₂) = 1.12 ÷ 24.0 = 0.04667 mol Mole ratio M : H₂ = 1 : 1 n(M) = 0.04667 mol Aᵣ(M) = mass ÷ moles = 2.00 ÷ 0.04667 = 42.9 Metal is calcium, Ca (Aᵣ = 40.1) ✓ (Note: 42.9 is closest to Ca; accept Ca with reasoning.)

Marking:

  • 1 mark: correct moles of H₂
  • 1 mark: correct Aᵣ calculation
  • 1 mark: correct identification of Ca

19. PₓClᵧ: 14.9% P; 0.200 g occupies 35.5 cm³ at 150 °C, 100 kPa. Find molecular formula. [3 marks]

Answer: Mass of Cl = 100 − 14.9 = 85.1%

Element%÷ AᵣRatio
P14.914.9 ÷ 31.0 = 0.4810.481 ÷ 0.481 = 1
Cl85.185.1 ÷ 35.5 = 2.402.40 ÷ 0.481 = 5
Empirical formula = PCl₅

Using ideal gas equation: p = 100 kPa = 100000 Pa; V = 35.5 cm³ = 3.55 × 10⁻⁵ m³; T = 150 + 273 = 423 K n = pV ÷ RT = (100000 × 3.55 × 10⁻⁵) ÷ (8.31 × 423) = 0.00101 mol Mᵣ = 0.200 ÷ 0.00101 = 198 Mᵣ(PCl₅) = 31.0 + (5 × 35.5) = 208.5 Ratio = 198 ÷ 208.5 ≈ 1 Molecular formula = PCl₅ ✓

Marking:

  • 1 mark: correct empirical formula PCl₅
  • 1 mark: correct Mᵣ from ideal gas equation
  • 1 mark: correct molecular formula PCl₅

20. 1.00 g Al/Zn alloy + excess HCl → 1.20 dm³ H₂ at 25 °C, 100 kPa (water vapour pressure 3.17 kPa). Calculate % Al. [4 marks]

Answer: Pressure of dry H₂ = 100 − 3.17 = 96.83 kPa = 96830 Pa V = 1.20 dm³ = 1.20 × 10⁻³ m³; T = 25 + 273 = 298 K n(H₂) total = pV ÷ RT = (96830 × 1.20 × 10⁻³) ÷ (8.31 × 298) = 0.0469 mol

Let mass of Al = x g; mass of Zn = (1.00 − x) g 2Al + 6HCl → 2AlCl₃ + 3H₂: n(H₂) from Al = (x ÷ 27.0) × (3/2) = 0.05556x Zn + 2HCl → ZnCl₂ + H₂: n(H₂) from Zn = (1.00 − x) ÷ 65.4 = 0.01529(1.00 − x)

Total n(H₂) = 0.05556x + 0.01529(1.00 − x) = 0.0469 0.05556x + 0.01529 − 0.01529x = 0.0469 0.04027x = 0.03161 x = 0.785 g

% Al = (0.785 ÷ 1.00) × 100 = 78.5% ✓

Marking:

  • 1 mark: correct pressure correction and moles of H₂
  • 1 mark: correct algebraic setup with mole ratios
  • 1 mark: correct solving for x
  • 1 mark: correct percentage

END OF ANSWER KEY