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A Level H1 Chemistry Stoichiometry Moles Quiz

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A-Level Chemistry H1 Quiz - Stoichiometry Moles

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 40

Duration: 45 minutes Total Marks: 40

Instructions:

  • Answer ALL questions in the spaces provided.
  • Show all working clearly; marks are awarded for method.
  • Use appropriate units and significant figures throughout.
  • A Periodic Table and Data Booklet may be used.
  • [Ar values: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Na = 23.0, S = 32.1, Cl = 35.5, Ca = 40.1, Fe = 55.8, Cu = 63.5]

Section A: Short Answer and Basic Calculations (10 marks)

Answer all questions in this section.

1. Define the term mole. [1]

2. Calculate the relative molecular mass, Mr, of ammonium sulfate, (NH₄)₂SO₄. [1]

3. Calculate the number of moles of sodium hydroxide, NaOH, present in 8.00 g of the solid. [1]

4. Calculate the mass of carbon dioxide, CO₂, produced when 0.500 mol of propane, C₃H₈, undergoes complete combustion. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O [2]

5. A sample of hydrated copper(II) sulfate, CuSO₄·xH₂O, has a mass of 4.99 g. After heating to constant mass, the anhydrous CuSO₄ has a mass of 3.19 g. Calculate the value of x. [3]

Section B: Stoichiometric Calculations (10 marks)

Answer all questions in this section.

6. Calculate the volume occupied by 0.250 mol of nitrogen gas, N₂, at room temperature and pressure (RTP). [Molar volume at RTP = 24.0 dm³ mol⁻¹] [1]

7. Calculate the concentration, in mol dm⁻³, of a solution prepared by dissolving 5.85 g of sodium chloride, NaCl, in water and making up to 250 cm³. [1]

8. 2.70 g of aluminium reacts completely with excess dilute hydrochloric acid according to the equation: 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)

(a) Calculate the amount, in moles, of aluminium used. [1]

(b) Calculate the amount, in moles, of hydrogen gas produced. [1]

(c) Calculate the volume of hydrogen gas produced at RTP. [1]

9. 25.0 cm³ of a solution of sulfuric acid, H₂SO₄, required 20.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide, NaOH, for complete neutralisation. 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

(a) Calculate the amount, in moles, of NaOH used. [1]

(b) Calculate the amount, in moles, of H₂SO₄ in 25.0 cm³ of the acid solution. [1]

(c) Calculate the concentration of the sulfuric acid in mol dm⁻³. [1]

10. A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

(a) Calculate the empirical formula of the compound. [3]

(b) The relative molecular mass of the compound is 180. Determine its molecular formula. [1]

Section C: Limiting Reagents and Applied Stoichiometry (10 marks)

Answer all questions in this section.

11. 1.20 g of magnesium reacts with 50.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid. Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

(a) Determine which reactant is the limiting reagent. Show your working. [3]

(b) Calculate the mass of magnesium chloride, MgCl₂, that can be formed. [2]

(c) Calculate the volume of hydrogen gas produced at RTP. [1]

12. A student carried out an experiment to determine the concentration of a solution of ethanoic acid, CH₃COOH, by titration with 0.100 mol dm⁻³ sodium hydroxide solution. 25.0 cm³ of the acid was pipetted into a conical flask and titrated. The following results were obtained:

Titration123
Final burette reading / cm³24.5047.8023.60
Initial burette reading / cm³0.0023.400.00
Volume of NaOH used / cm³24.5024.4023.60

(a) Which titrations should be used to calculate the average volume of NaOH? Explain your answer. [2]

(b) Calculate the average volume of NaOH used. [1]

(c) Calculate the concentration of the ethanoic acid solution in mol dm⁻³. CH₃COOH + NaOH → CH₃COONa + H₂O [2]

13. A 0.500 g sample of an impure mixture containing sodium hydrogencarbonate, NaHCO₃, was heated strongly. The NaHCO₃ decomposed according to the equation: 2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g)

The mass of the residue after heating was 0.318 g.

(a) Calculate the mass of NaHCO₃ that decomposed. [3]

(b) Calculate the percentage purity of NaHCO₃ in the original sample. [2]

14. A fertiliser contains ammonium nitrate, NH₄NO₃, as its only nitrogen-containing compound.

(a) Calculate the percentage by mass of nitrogen in pure NH₄NO₃. [2]

(b) A 5.00 g sample of the fertiliser was analysed and found to contain 1.40 g of nitrogen. Calculate the percentage purity of NH₄NO₃ in the fertiliser. [2]

15. Calculate the mass of calcium carbonate, CaCO₃, required to produce 1.20 dm³ of carbon dioxide gas at RTP when reacted with excess hydrochloric acid. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) [2]

Section D: Data Interpretation and Advanced Stoichiometry (10 marks)

Answer all questions in this section.

16. A 2.50 g sample of an iron ore was dissolved in acid and all the iron was reduced to Fe²⁺ ions. The solution was titrated with 0.0200 mol dm⁻³ potassium manganate(VII), KMnO₄, in acidic conditions. 24.0 cm³ of KMnO₄ solution was required to reach the endpoint. MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

(a) Calculate the amount, in moles, of MnO₄⁻ used in the titration. [1]

(b) Calculate the amount, in moles, of Fe²⁺ in the solution. [1]

(c) Calculate the mass of iron in the ore sample. [1]

(d) Calculate the percentage by mass of iron in the ore. [1]

17. A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 56.0.

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

(b) Determine the molecular formula of the hydrocarbon. [1]

18. 10.0 cm³ of a solution of sodium carbonate, Na₂CO₃, reacts completely with 15.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Calculate the concentration of the sodium carbonate solution in mol dm⁻³. [2]

19. A student prepared a standard solution by dissolving 2.65 g of anhydrous sodium carbonate, Na₂CO₃, in water and making up to 250 cm³. 25.0 cm³ of this solution required 20.0 cm³ of hydrochloric acid for complete neutralisation.

(a) Calculate the concentration of the sodium carbonate standard solution in mol dm⁻³. [2]

(b) Calculate the concentration of the hydrochloric acid in mol dm⁻³. [2]

20. A 1.00 g sample of a mixture of sodium chloride and sodium nitrate 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 silver chloride obtained was 1.44 g. Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

(a) Calculate the amount, in moles, of silver chloride formed. [1]

(b) Calculate the mass of sodium chloride in the original mixture. [2]

(c) Calculate the percentage by mass of sodium nitrate in the mixture. [1]

END OF QUIZ

Check your work carefully. Ensure all answers include appropriate units and significant figures.

Answers

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A-Level Chemistry H1 Quiz - Stoichiometry Moles: Answer Key

Total Marks: 40

Section A: Short Answer and Basic Calculations (10 marks)

1. Define the term mole.

  • Answer: A mole is the amount of substance that contains the same number of particles (atoms, molecules, ions, or other entities) as there are atoms in exactly 12.0 g of carbon-12. [1]
  • Marking note: Accept "Avogadro's number of particles (6.02 × 10²³)" or equivalent definition. Must reference number of particles or Avogadro's constant.

2. Calculate the relative molecular mass, Mr, of ammonium sulfate, (NH₄)₂SO₄.

  • Answer: Mr = 2[14.0 + 4(1.0)] + 32.1 + 4(16.0) = 2(18.0) + 32.1 + 64.0 = 36.0 + 32.1 + 64.0 = 132.1 [1]
  • Marking note: Award mark for correct answer. Allow 132 if rounded.

3. Calculate the number of moles of sodium hydroxide, NaOH, present in 8.00 g of the solid.

  • Answer: Mr(NaOH) = 23.0 + 16.0 + 1.0 = 40.0; n = m/Mr = 8.00/40.0 = 0.200 mol [1]
  • Marking note: Award mark for correct answer with or without working shown.

4. Calculate the mass of carbon dioxide, CO₂, produced when 0.500 mol of propane, C₃H₈, undergoes complete combustion. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

  • Answer: Mole ratio C₃H₈ : CO₂ = 1 : 3 [1]; n(CO₂) = 3 × 0.500 = 1.50 mol; Mr(CO₂) = 44.0; mass = 1.50 × 44.0 = 66.0 g [1]
  • Marking note: 1 mark for correct mole ratio/calculation of moles of CO₂; 1 mark for correct mass.

5. A sample of hydrated copper(II) sulfate, CuSO₄·xH₂O, has a mass of 4.99 g. After heating to constant mass, the anhydrous CuSO₄ has a mass of 3.19 g. Calculate the value of x.

  • Answer: Mass of water lost = 4.99 – 3.19 = 1.80 g [1]; Mr(CuSO₄) = 63.5 + 32.1 + 64.0 = 159.6; n(CuSO₄) = 3.19/159.6 = 0.0200 mol [1]; n(H₂O) = 1.80/18.0 = 0.100 mol; Ratio H₂O : CuSO₄ = 0.100/0.0200 = 5; x = 5 [1]
  • Marking note: 1 mark for mass of water; 1 mark for moles of CuSO₄ or H₂O; 1 mark for correct x value.

Section B: Stoichiometric Calculations (10 marks)

6. Calculate the volume occupied by 0.250 mol of nitrogen gas, N₂, at room temperature and pressure (RTP).

  • Answer: Volume = n × 24.0 = 0.250 × 24.0 = 6.00 dm³ [1]
  • Marking note: Accept 6.0 dm³. Unit must be stated or implied.

7. Calculate the concentration, in mol dm⁻³, of a solution prepared by dissolving 5.85 g of sodium chloride, NaCl, in water and making up to 250 cm³.

  • Answer: Mr(NaCl) = 23.0 + 35.5 = 58.5; n = 5.85/58.5 = 0.100 mol; c = n/V = 0.100/(250/1000) = 0.400 mol dm⁻³ [1]
  • Marking note: Award mark for correct final answer.

8. 2.70 g of aluminium reacts completely with excess dilute hydrochloric acid. 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)

(a) Calculate the amount, in moles, of aluminium used.

  • Answer: n(Al) = 2.70/27.0 = 0.100 mol [1]

(b) Calculate the amount, in moles, of hydrogen gas produced.

  • Answer: Mole ratio Al : H₂ = 2 : 3; n(H₂) = 0.100 × (3/2) = 0.150 mol [1]

(c) Calculate the volume of hydrogen gas produced at RTP.

  • Answer: Volume = 0.150 × 24.0 = 3.60 dm³ [1]

9. 25.0 cm³ of sulfuric acid required 20.0 cm³ of 0.100 mol dm⁻³ NaOH for neutralisation. 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

(a) Calculate the amount, in moles, of NaOH used.

  • Answer: n(NaOH) = 0.100 × (20.0/1000) = 0.00200 mol [1]

(b) Calculate the amount, in moles, of H₂SO₄ in 25.0 cm³ of the acid solution.

  • Answer: Mole ratio NaOH : H₂SO₄ = 2 : 1; n(H₂SO₄) = 0.00200/2 = 0.00100 mol [1]

(c) Calculate the concentration of the sulfuric acid in mol dm⁻³.

  • Answer: c = n/V = 0.00100/(25.0/1000) = 0.0400 mol dm⁻³ [1]

10. A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

(a) Calculate the empirical formula of the compound.

  • Answer: Assume 100 g sample: C = 40.0 g, H = 6.7 g, O = 53.3 g [1]; Moles: C = 40.0/12.0 = 3.33, H = 6.7/1.0 = 6.7, O = 53.3/16.0 = 3.33 [1]; Divide by smallest (3.33): C = 1, H = 2, O = 1; Empirical formula = CH₂O [1]
  • Marking note: 1 mark for correct mole calculations; 1 mark for correct ratio; 1 mark for correct formula.

(b) The relative molecular mass of the compound is 180. Determine its molecular formula.

  • Answer: Mr(CH₂O) = 12.0 + 2.0 + 16.0 = 30.0; n = 180/30.0 = 6; Molecular formula = C₆H₁₂O₆ [1]

Section C: Limiting Reagents and Applied Stoichiometry (10 marks)

11. 1.20 g of magnesium reacts with 50.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid. Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

(a) Determine which reactant is the limiting reagent. Show your working.

  • Answer: n(Mg) = 1.20/24.3 = 0.0494 mol [1]; n(HCl) = 2.00 × (50.0/1000) = 0.100 mol [1]; From equation, 1 mol Mg requires 2 mol HCl. 0.0494 mol Mg requires 0.0988 mol HCl. Available HCl (0.100 mol) > required (0.0988 mol), so Mg is the limiting reagent. [1]
  • Marking note: 1 mark each for moles of Mg and HCl; 1 mark for correct identification with reasoning. Accept alternative method comparing mole ratios.

(b) Calculate the mass of magnesium chloride, MgCl₂, that can be formed.

  • Answer: n(MgCl₂) = n(Mg) = 0.0494 mol [1]; Mr(MgCl₂) = 24.3 + 2(35.5) = 95.3; mass = 0.0494 × 95.3 = 4.71 g [1]
  • Marking note: 1 mark for correct moles of MgCl₂; 1 mark for correct mass.

(c) Calculate the volume of hydrogen gas produced at RTP.

  • Answer: n(H₂) = n(Mg) = 0.0494 mol; Volume = 0.0494 × 24.0 = 1.19 dm³ [1]

12. Titration of ethanoic acid with 0.100 mol dm⁻³ NaOH.

(a) Which titrations should be used to calculate the average volume of NaOH? Explain your answer.

  • Answer: Titrations 1 and 2 should be used. [1] Titration 3 (23.60 cm³) is not concordant with titrations 1 and 2 (24.50 and 24.40 cm³). Concordant results should agree within ±0.10 cm³. Titration 3 differs by more than 0.10 cm³ from the others and is likely an outlier/rough titration. [1]
  • Marking note: 1 mark for identifying titrations 1 and 2; 1 mark for valid explanation referencing concordancy.

(b) Calculate the average volume of NaOH used.

  • Answer: Average = (24.50 + 24.40)/2 = 24.45 cm³ [1]

(c) Calculate the concentration of the ethanoic acid solution in mol dm⁻³.

  • Answer: n(NaOH) = 0.100 × (24.45/1000) = 0.002445 mol [1]; n(CH₃COOH) = n(NaOH) = 0.002445 mol (1:1 ratio); c = 0.002445/(25.0/1000) = 0.0978 mol dm⁻³ [1]
  • Marking note: 1 mark for correct moles of NaOH; 1 mark for correct concentration. Accept 0.098 mol dm⁻³.

13. Decomposition of impure NaHCO₃.

(a) Calculate the mass of NaHCO₃ that decomposed.

  • Answer: Mass loss = mass of H₂O + CO₂ produced = 0.500 – 0.318 = 0.182 g [1]; Mr(H₂O + CO₂) = 18.0 + 44.0 = 62.0; n(H₂O + CO₂) = 0.182/62.0 = 0.002935 mol [1]; From equation, 2 mol NaHCO₃ produces 1 mol H₂O + 1 mol CO₂ (2 mol gases total). n(NaHCO₃) = n(gases) = 0.002935 mol; Mr(NaHCO₃) = 84.0; mass NaHCO₃ = 0.002935 × 84.0 = 0.247 g [1]
  • Marking note: 1 mark for mass loss; 1 mark for moles of gases; 1 mark for mass of NaHCO₃. Accept alternative method using mass ratio.

(b) Calculate the percentage purity of NaHCO₃ in the original sample.

  • Answer: Percentage purity = (0.247/0.500) × 100 = 49.4% [1] (Allow 49–50% depending on rounding)
  • Marking note: 1 mark for correct percentage; 1 mark for using answer from (a). Error carried forward (ECF) applies.

14. Fertiliser containing NH₄NO₃.

(a) Calculate the percentage by mass of nitrogen in pure NH₄NO₃.

  • Answer: Mr(NH₄NO₃) = 14.0 + 4.0 + 14.0 + 48.0 = 80.0 [1]; Mass of N = 2 × 14.0 = 28.0; Percentage N = (28.0/80.0) × 100 = 35.0% [1]
  • Marking note: 1 mark for correct Mr; 1 mark for correct percentage.

(b) A 5.00 g sample of the fertiliser was analysed and found to contain 1.40 g of nitrogen. Calculate the percentage purity of NH₄NO₃ in the fertiliser.

  • Answer: Mass of NH₄NO₃ needed to provide 1.40 g N = 1.40 × (80.0/28.0) = 4.00 g [1]; Percentage purity = (4.00/5.00) × 100 = 80.0% [1]
  • Marking note: 1 mark for mass of NH₄NO₃; 1 mark for correct percentage.

15. Calculate the mass of calcium carbonate, CaCO₃, required to produce 1.20 dm³ of carbon dioxide gas at RTP when reacted with excess hydrochloric acid. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

  • Answer: n(CO₂) = 1.20/24.0 = 0.0500 mol [1]; Mole ratio CaCO₃ : CO₂ = 1 : 1; n(CaCO₃) = 0.0500 mol; Mr(CaCO₃) = 40.1 + 12.0 + 48.0 = 100.1; mass = 0.0500 × 100.1 = 5.01 g [1]
  • Marking note: 1 mark for moles of CO₂; 1 mark for correct mass.

Section D: Data Interpretation and Advanced Stoichiometry (10 marks)

16. Iron ore analysis by redox titration.

(a) Calculate the amount, in moles, of MnO₄⁻ used in the titration.

  • Answer: n(MnO₄⁻) = 0.0200 × (24.0/1000) = 0.000480 mol [1]

(b) Calculate the amount, in moles, of Fe²⁺ in the solution.

  • Answer: Mole ratio MnO₄⁻ : Fe²⁺ = 1 : 5; n(Fe²⁺) = 5 × 0.000480 = 0.00240 mol [1]

(c) Calculate the mass of iron in the ore sample.

  • Answer: m(Fe) = n × Ar = 0.00240 × 55.8 = 0.134 g [1]

(d) Calculate the percentage by mass of iron in the ore.

  • Answer: Percentage Fe = (0.134/2.50) × 100 = 5.36% [1]

17. A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 56.0.

(a) Determine the empirical formula of the hydrocarbon.

  • Answer: Assume 100 g: C = 85.7 g, H = 14.3 g; Moles: C = 85.7/12.0 = 7.14, H = 14.3/1.0 = 14.3 [1]; Divide by smallest (7.14): C = 1, H = 2; Empirical formula = CH₂ [1]

(b) Determine the molecular formula of the hydrocarbon.

  • Answer: Mr(CH₂) = 14.0; n = 56.0/14.0 = 4; Molecular formula = C₄H₈ [1]

18. Calculate the concentration of sodium carbonate solution. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

  • Answer: n(HCl) = 0.500 × (15.0/1000) = 0.00750 mol [1]; Mole ratio Na₂CO₃ : HCl = 1 : 2; n(Na₂CO₃) = 0.00750/2 = 0.00375 mol; c = n/V = 0.00375/(10.0/1000) = 0.375 mol dm⁻³ [1]

19. Standard solution and titration.

(a) Calculate the concentration of the sodium carbonate standard solution in mol dm⁻³.

  • Answer: Mr(Na₂CO₃) = 2(23.0) + 12.0 + 3(16.0) = 106.0; n = 2.65/106.0 = 0.0250 mol [1]; c = 0.0250/(250/1000) = 0.100 mol dm⁻³ [1]

(b) Calculate the concentration of the hydrochloric acid in mol dm⁻³.

  • Answer: n(Na₂CO₃) in 25.0 cm³ = 0.100 × (25.0/1000) = 0.00250 mol [1]; Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂; n(HCl) = 2 × 0.00250 = 0.00500 mol; c(HCl) = 0.00500/(20.0/1000) = 0.250 mol dm⁻³ [1]

20. Gravimetric analysis of chloride mixture.

(a) Calculate the amount, in moles, of silver chloride formed.

  • Answer: Mr(AgCl) = 107.9 + 35.5 = 143.4; n(AgCl) = 1.44/143.4 = 0.0100 mol [1]

(b) Calculate the mass of sodium chloride in the original mixture.

  • Answer: n(Cl⁻) = n(AgCl) = 0.0100 mol; n(NaCl) = 0.0100 mol [1]; Mr(NaCl) = 58.5; mass NaCl = 0.0100 × 58.5 = 0.585 g [1]

(c) Calculate the percentage by mass of sodium nitrate in the mixture.

  • Answer: Mass of NaNO₃ = 1.00 – 0.585 = 0.415 g; Percentage NaNO₃ = (0.415/1.00) × 100 = 41.5% [1]

END OF ANSWER KEY