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Secondary 4 Combined Science Chemistry Stoichiometry Moles Quiz

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Questions

Secondary 4 Combined Science Chemistry Quiz - Stoichiometry Moles

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

Duration: 45 minutes Total Marks: 40

Instructions:

This quiz contains 20 questions on Stoichiometry and the Mole Concept.
Answer ALL questions in the spaces provided.
Show all working for calculation questions. Marks are awarded for method.
Relative atomic masses: H=1, C=12, N=14, O=16, Na=23, Mg=24, Al=27, S=32, Cl=35.5, Ca=40, Fe=56, Cu=63.5, Zn=65
Molar volume of gas at RTP = 24 dm³

Section A: Short Answer and Calculation (Questions 1–10)

Answer all questions. Marks are indicated in brackets.

1. Define the term "mole" in chemistry. (2 marks)

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

3. Calculate the number of moles in 8.5 g of sodium nitrate, NaNO₃. (2 marks)

4. A student dissolves 5.85 g of sodium chloride, NaCl, in water to make 250 cm³ of solution. Calculate the concentration of the solution in mol/dm³. (3 marks)

5. Magnesium reacts with hydrochloric acid according to the equation: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Calculate the mass of magnesium needed to produce 4.8 dm³ of hydrogen gas at RTP. (3 marks)

6. Calculate the percentage by mass of nitrogen in ammonium nitrate, NH₄NO₃. (2 marks)

7. Iron(III) oxide reacts with carbon monoxide according to the equation: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

Calculate the mass of iron produced when 16.0 g of iron(III) oxide reacts completely. (3 marks)

8. A solution of sodium hydroxide has a concentration of 0.50 mol/dm³. What volume of this solution contains 0.025 moles of sodium hydroxide? Give your answer in cm³. (2 marks)

9. Calculate the number of molecules in 0.20 moles of carbon dioxide, CO₂. (Avogadro constant = 6.0 × 10²³) (1 mark)

10. A compound has the empirical formula CH₂O and a relative molecular mass of 180. Determine its molecular formula. (2 marks)

Section B: Structured Questions (Questions 11–15)

Answer all questions. Marks are indicated in brackets.

11. A student carries out a titration to determine the concentration of a solution of sulfuric acid, H₂SO₄.

The student uses 25.0 cm³ of sulfuric acid and finds that 20.0 cm³ of 0.50 mol/dm³ sodium hydroxide, NaOH, is needed for complete neutralisation.

The equation for the reaction is: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

(a) Calculate the number of moles of NaOH used in the titration. (1 mark)

(b) Using the equation, state the mole ratio of H₂SO₄ to NaOH. (1 mark)

(c) Calculate the number of moles of H₂SO₄ in 25.0 cm³ of the acid. (1 mark)

(d) Calculate the concentration of the sulfuric acid in mol/dm³. (2 marks)

12. Calcium carbonate, CaCO₃, decomposes on heating: CaCO₃(s) → CaO(s) + CO₂(g)

(a) Calculate the relative formula mass of calcium carbonate. (1 mark)

(b) Calculate the mass of calcium oxide, CaO, produced when 25.0 g of calcium carbonate is heated completely. (2 marks)

(c) Calculate the volume of carbon dioxide gas produced at RTP from the decomposition in part (b). (2 marks)

13. A student prepared a standard solution by dissolving 10.6 g of anhydrous sodium carbonate, Na₂CO₃, in distilled water and making the solution up to 500 cm³.

(a) Calculate the relative formula mass of Na₂CO₃. (1 mark)

(b) Calculate the number of moles of Na₂CO₃ dissolved. (1 mark)

(c) Calculate the concentration of the sodium carbonate solution in mol/dm³. (1 mark)

(d) The student then takes 25.0 cm³ of this solution and dilutes it to 250 cm³. Calculate the concentration of the diluted solution. (2 marks)

14. Propane, C₃H₈, burns completely in oxygen: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

(a) Calculate the relative molecular mass of propane. (1 mark)

(b) Calculate the number of moles in 4.4 g of propane. (1 mark)

(c) Using the equation, calculate the volume of oxygen gas needed for the complete combustion of 4.4 g of propane at RTP. (2 marks)

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

(a) Calculate the empirical formula of the compound. (3 marks)

(b) The relative molecular mass of the compound is 60. Determine its molecular formula. (1 mark)

Section C: Data-Based and Application Questions (Questions 16–20)

Answer all questions. Marks are indicated in brackets.

16. The table below shows the mass of four samples of different substances and their relative formula masses.

Substance	Mass (g)	Relative formula mass (Mᵣ)
W	4.0	40
X	7.1	71
Y	8.8	44
Z	11.7	58.5

(a) Which substance contains the largest number of moles? Show your reasoning. (2 marks)

(b) Which substance contains the smallest number of moles? Show your reasoning. (1 mark)

17. A student investigated the reaction between magnesium ribbon and excess dilute hydrochloric acid. The student measured the volume of hydrogen gas produced over time.

The equation for the reaction is: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

The student used 0.12 g of magnesium ribbon.

(a) Calculate the number of moles of magnesium used. (1 mark)

(b) Calculate the theoretical volume of hydrogen gas that should be produced at RTP. (2 marks)

(c) The student collected only 100 cm³ of hydrogen gas. Calculate the percentage yield of hydrogen. (2 marks)

18. A factory produces ammonia, NH₃, by reacting nitrogen gas with hydrogen gas: N₂(g) + 3H₂(g) → 2NH₃(g)

(a) Calculate the relative molecular mass of ammonia. (1 mark)

(b) The factory uses 280 kg of nitrogen gas. Calculate the number of moles of nitrogen used. (1 kg = 1000 g) (2 marks)

(c) Using the equation, calculate the maximum mass of ammonia that can be produced from 280 kg of nitrogen. (2 marks)

19. A student dissolved an unknown mass of potassium hydroxide, KOH, in water to make 100 cm³ of solution. The student then titrated 25.0 cm³ of this solution against 0.20 mol/dm³ hydrochloric acid, HCl.

The equation for the reaction is: KOH(aq) + HCl(aq) → KCl(aq) + H₂O(l)

The student found that 15.0 cm³ of the acid was needed to neutralise 25.0 cm³ of the potassium hydroxide solution.

(a) Calculate the number of moles of HCl used in the titration. (1 mark)

(b) Calculate the concentration of the potassium hydroxide solution in mol/dm³. (2 marks)

(c) Calculate the mass of potassium hydroxide the student originally dissolved in 100 cm³ of water. (2 marks)

20. Ethene, C₂H₄, reacts with steam to form ethanol, C₂H₅OH: C₂H₄(g) + H₂O(g) → C₂H₅OH(l)

(a) Calculate the relative molecular mass of ethanol. (1 mark)

(b) In an industrial process, 14.0 tonnes of ethene produced 18.4 tonnes of ethanol. (1 tonne = 1 × 10⁶ g)

(i) Calculate the theoretical yield of ethanol from 14.0 tonnes of ethene. (2 marks)

(ii) Calculate the percentage yield of ethanol in this process. (2 marks)

END OF QUIZ

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Answers

Secondary 4 Combined Science Chemistry Quiz - Stoichiometry Moles

ANSWER KEY AND MARKING SCHEME

Total Marks: 40

Section A: Short Answer and Calculation (Questions 1–10)

1. Define the term "mole" in chemistry. (2 marks)

Answer: A mole is the amount of substance that contains the same number of particles (atoms, molecules, or ions) as there are atoms in exactly 12 g of carbon-12. [1 mark] This number is the Avogadro constant, 6.0 × 10²³. [1 mark]

Marking notes: Accept "amount of substance containing 6.0 × 10²³ particles" for both marks. Award 1 mark for mentioning Avogadro's number/constant, 1 mark for linking to amount of substance or number of particles.

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

Answer:

N: 2 × 14 = 28
H: 8 × 1 = 8
S: 1 × 32 = 32
O: 4 × 16 = 64
Mᵣ = 28 + 8 + 32 + 64 = 132 [2 marks]

Marking notes: Award 1 mark for correct identification of atoms and multiplication by subscripts. Award 2 marks for correct final answer of 132. Accept working showing step-by-step addition.

3. Calculate the number of moles in 8.5 g of sodium nitrate, NaNO₃. (2 marks)

Answer:

Mᵣ of NaNO₃ = 23 + 14 + (3 × 16) = 23 + 14 + 48 = 85 [1 mark]
Number of moles = mass / Mᵣ = 8.5 / 85 = 0.10 mol [1 mark]

Marking notes: Award 1 mark for correct Mᵣ calculation. Award 1 mark for correct mole calculation with units. Accept 0.1 mol.

4. A student dissolves 5.85 g of sodium chloride, NaCl, in water to make 250 cm³ of solution. Calculate the concentration of the solution in mol/dm³. (3 marks)

Answer:

Mᵣ of NaCl = 23 + 35.5 = 58.5 [1 mark]
Moles of NaCl = 5.85 / 58.5 = 0.10 mol [1 mark]
Volume in dm³ = 250 / 1000 = 0.250 dm³
Concentration = moles / volume = 0.10 / 0.250 = 0.40 mol/dm³ [1 mark]

Marking notes: Award 1 mark for Mᵣ, 1 mark for moles calculation, 1 mark for correct concentration with units. Deduct 1 mark if volume conversion is omitted but final answer is numerically correct by coincidence.

5. Magnesium reacts with hydrochloric acid according to the equation: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Calculate the mass of magnesium needed to produce 4.8 dm³ of hydrogen gas at RTP. (3 marks)

Answer:

Moles of H₂ = volume / molar volume = 4.8 / 24 = 0.20 mol [1 mark]
From equation: 1 mol Mg produces 1 mol H₂, so moles of Mg needed = 0.20 mol [1 mark]
Mass of Mg = moles × Aᵣ = 0.20 × 24 = 4.8 g [1 mark]

Marking notes: Award 1 mark for moles of H₂, 1 mark for mole ratio application, 1 mark for mass calculation with correct units. Accept alternative correct methods.

6. Calculate the percentage by mass of nitrogen in ammonium nitrate, NH₄NO₃. (2 marks)

Answer:

Mᵣ of NH₄NO₃ = 14 + (4 × 1) + 14 + (3 × 16) = 14 + 4 + 14 + 48 = 80 [1 mark]
Mass of nitrogen = 2 × 14 = 28
Percentage of N = (28 / 80) × 100 = 35% [1 mark]

Marking notes: Award 1 mark for correct Mᵣ. Award 1 mark for correct percentage calculation. Accept 35.0%.

7. Iron(III) oxide reacts with carbon monoxide according to the equation: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

Calculate the mass of iron produced when 16.0 g of iron(III) oxide reacts completely. (3 marks)

Answer:

Mᵣ of Fe₂O₃ = (2 × 56) + (3 × 16) = 112 + 48 = 160 [1 mark]
Moles of Fe₂O₃ = 16.0 / 160 = 0.10 mol [1 mark]
From equation: 1 mol Fe₂O₃ produces 2 mol Fe
Moles of Fe = 2 × 0.10 = 0.20 mol
Mass of Fe = 0.20 × 56 = 11.2 g [1 mark]

Marking notes: Award 1 mark for Mᵣ of Fe₂O₃, 1 mark for moles of Fe₂O₃, 1 mark for final mass with units. Accept 11.2 g.

8. A solution of sodium hydroxide has a concentration of 0.50 mol/dm³. What volume of this solution contains 0.025 moles of sodium hydroxide? Give your answer in cm³. (2 marks)

Answer:

Volume (dm³) = moles / concentration = 0.025 / 0.50 = 0.050 dm³ [1 mark]
Volume (cm³) = 0.050 × 1000 = 50 cm³ [1 mark]

Marking notes: Award 1 mark for correct volume in dm³, 1 mark for conversion to cm³. Accept 50 cm³.

9. Calculate the number of molecules in 0.20 moles of carbon dioxide, CO₂. (Avogadro constant = 6.0 × 10²³) (1 mark)

Answer:

Number of molecules = 0.20 × 6.0 × 10²³ = 1.2 × 10²³ [1 mark]

Marking notes: Award 1 mark for correct answer. Accept 1.20 × 10²³.

10. 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 + (2 × 1) + 16 = 30 [1 mark]
n = Mᵣ of compound / Mᵣ of empirical formula = 180 / 30 = 6
Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ [1 mark]

Marking notes: Award 1 mark for calculating empirical formula mass and determining n. Award 1 mark for correct molecular formula.

Section B: Structured Questions (Questions 11–15)

11. Titration of sulfuric acid with sodium hydroxide.

(a) Calculate the number of moles of NaOH used in the titration. (1 mark)

Answer:

Moles of NaOH = concentration × volume (dm³) = 0.50 × (20.0 / 1000) = 0.50 × 0.020 = 0.010 mol [1 mark]

Marking notes: Award 1 mark for correct answer with or without units. Accept 0.01 mol.

(b) Using the equation, state the mole ratio of H₂SO₄ to NaOH. (1 mark)

Answer: H₂SO₄ : NaOH = 1 : 2 [1 mark]

Marking notes: Must be expressed as a ratio. Accept "1 mole of H₂SO₄ reacts with 2 moles of NaOH."

(c) Calculate the number of moles of H₂SO₄ in 25.0 cm³ of the acid. (1 mark)

Answer:

Moles of H₂SO₄ = moles of NaOH / 2 = 0.010 / 2 = 0.0050 mol [1 mark]

Marking notes: Award 1 mark for correct answer. Accept 0.005 mol.

(d) Calculate the concentration of the sulfuric acid in mol/dm³. (2 marks)

Answer:

Volume of H₂SO₄ in dm³ = 25.0 / 1000 = 0.0250 dm³ [1 mark]
Concentration = moles / volume = 0.0050 / 0.0250 = 0.20 mol/dm³ [1 mark]

Marking notes: Award 1 mark for volume conversion, 1 mark for correct concentration with units.

12. Decomposition of calcium carbonate.

(a) Calculate the relative formula mass of calcium carbonate. (1 mark)

Answer:

Mᵣ of CaCO₃ = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100 [1 mark]

Marking notes: Award 1 mark for correct answer.

(b) Calculate the mass of calcium oxide, CaO, produced when 25.0 g of calcium carbonate is heated completely. (2 marks)

Answer:

Moles of CaCO₃ = 25.0 / 100 = 0.250 mol [1 mark]
From equation: 1 mol CaCO₃ produces 1 mol CaO
Moles of CaO = 0.250 mol
Mᵣ of CaO = 40 + 16 = 56
Mass of CaO = 0.250 × 56 = 14.0 g [1 mark]

Marking notes: Award 1 mark for moles of CaCO₃, 1 mark for final mass with units.

(c) Calculate the volume of carbon dioxide gas produced at RTP from the decomposition in part (b). (2 marks)

Answer:

From equation: 1 mol CaCO₃ produces 1 mol CO₂
Moles of CO₂ = 0.250 mol [1 mark]
Volume of CO₂ = 0.250 × 24 = 6.0 dm³ [1 mark]

Marking notes: Award 1 mark for moles of CO₂, 1 mark for volume with units. Accept 6 dm³.

13. Preparation and dilution of sodium carbonate solution.

(a) Calculate the relative formula mass of Na₂CO₃. (1 mark)

Answer:

Mᵣ of Na₂CO₃ = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106 [1 mark]

Marking notes: Award 1 mark for correct answer.

(b) Calculate the number of moles of Na₂CO₃ dissolved. (1 mark)

Answer:

Moles = 10.6 / 106 = 0.100 mol [1 mark]

Marking notes: Award 1 mark for correct answer. Accept 0.10 mol.

(c) Calculate the concentration of the sodium carbonate solution in mol/dm³. (1 mark)

Answer:

Volume = 500 / 1000 = 0.500 dm³
Concentration = 0.100 / 0.500 = 0.20 mol/dm³ [1 mark]

Marking notes: Award 1 mark for correct answer with units. Accept 0.2 mol/dm³.

(d) The student then takes 25.0 cm³ of this solution and dilutes it to 250 cm³. Calculate the concentration of the diluted solution. (2 marks)

Answer:

Moles in 25.0 cm³ = 0.20 × (25.0 / 1000) = 0.0050 mol [1 mark]
New concentration = 0.0050 / (250 / 1000) = 0.0050 / 0.250 = 0.020 mol/dm³ [1 mark]

Alternative method:

Dilution factor = 250 / 25.0 = 10
New concentration = 0.20 / 10 = 0.020 mol/dm³

Marking notes: Award 1 mark for correct method (either moles in aliquot or dilution factor), 1 mark for correct final concentration with units.

14. Combustion of propane.

(a) Calculate the relative molecular mass of propane. (1 mark)

Answer:

Mᵣ of C₃H₈ = (3 × 12) + (8 × 1) = 36 + 8 = 44 [1 mark]

Marking notes: Award 1 mark for correct answer.

(b) Calculate the number of moles in 4.4 g of propane. (1 mark)

Answer:

Moles = 4.4 / 44 = 0.10 mol [1 mark]

Marking notes: Award 1 mark for correct answer. Accept 0.1 mol.

(c) Using the equation, calculate the volume of oxygen gas needed for the complete combustion of 4.4 g of propane at RTP. (2 marks)

Answer:

From equation: 1 mol C₃H₈ reacts with 5 mol O₂
Moles of O₂ = 5 × 0.10 = 0.50 mol [1 mark]
Volume of O₂ = 0.50 × 24 = 12 dm³ [1 mark]

Marking notes: Award 1 mark for correct mole ratio application, 1 mark for volume with units.

15. Empirical and molecular formula determination.

(a) Calculate the empirical formula of the compound. (3 marks)

Answer:

Element	% by mass	Aᵣ	Moles = % / Aᵣ	Simplest ratio
C	40.0	12	40.0/12 = 3.33	3.33/3.33 = 1
H	6.7	1	6.7/1 = 6.7	6.7/3.33 = 2
O	53.3	16	53.3/16 = 3.33	3.33/3.33 = 1

[2 marks for correct table/method]

Empirical formula = CH₂O [1 mark]

Marking notes: Award 1 mark for correct mole calculations, 1 mark for correct ratio determination, 1 mark for empirical formula. Accept working without a table if clear.

(b) The relative molecular mass of the compound is 60. Determine its molecular formula. (1 mark)

Answer:

Mᵣ of CH₂O = 12 + 2 + 16 = 30
n = 60 / 30 = 2
Molecular formula = C₂H₄O₂ [1 mark]

Marking notes: Award 1 mark for correct molecular formula.

Section C: Data-Based and Application Questions (Questions 16–20)

16. Analysis of substance samples.

(a) Which substance contains the largest number of moles? Show your reasoning. (2 marks)

Answer:

Moles = mass / Mᵣ
- W: 4.0 / 40 = 0.10 mol
- X: 7.1 / 71 = 0.10 mol
- Y: 8.8 / 44 = 0.20 mol
- Z: 11.7 / 58.5 = 0.20 mol
Y and Z both contain 0.20 mol, which is the largest number of moles. [2 marks]

Marking notes: Award 1 mark for calculating moles for all substances, 1 mark for identifying Y and Z as having the largest number of moles. Accept either Y or Z if student states both are equal.

(b) Which substance contains the smallest number of moles? Show your reasoning. (1 mark)

Answer:

W and X both contain 0.10 mol, which is the smallest number of moles. [1 mark]

Marking notes: Award 1 mark for identifying W and X. Accept either W or X if student states both are equal.

17. Reaction of magnesium with hydrochloric acid.

(a) Calculate the number of moles of magnesium used. (1 mark)

Answer:

Moles of Mg = 0.12 / 24 = 0.0050 mol [1 mark]

Marking notes: Award 1 mark for correct answer. Accept 0.005 mol.

(b) Calculate the theoretical volume of hydrogen gas that should be produced at RTP. (2 marks)

Answer:

From equation: 1 mol Mg produces 1 mol H₂
Moles of H₂ = 0.0050 mol [1 mark]
Volume of H₂ = 0.0050 × 24 = 0.12 dm³ = 120 cm³ [1 mark]

Marking notes: Award 1 mark for moles of H₂, 1 mark for volume with units. Accept 120 cm³ or 0.12 dm³.

(c) The student collected only 100 cm³ of hydrogen gas. Calculate the percentage yield of hydrogen. (2 marks)

Answer:

Percentage yield = (actual yield / theoretical yield) × 100 [1 mark]
Percentage yield = (100 / 120) × 100 = 83.3% [1 mark]

Marking notes: Award 1 mark for correct formula, 1 mark for correct answer. Accept 83% or 83.3%.

18. Industrial production of ammonia.

(a) Calculate the relative molecular mass of ammonia. (1 mark)

Answer:

Mᵣ of NH₃ = 14 + (3 × 1) = 17 [1 mark]

Marking notes: Award 1 mark for correct answer.

(b) The factory uses 280 kg of nitrogen gas. Calculate the number of moles of nitrogen used. (1 kg = 1000 g) (2 marks)

Answer:

Mass of N₂ in grams = 280 × 1000 = 280,000 g [1 mark]
Mᵣ of N₂ = 2 × 14 = 28
Moles of N₂ = 280,000 / 28 = 10,000 mol [1 mark]

Marking notes: Award 1 mark for mass conversion, 1 mark for correct mole calculation. Accept 1.0 × 10⁴ mol.

(c) Using the equation, calculate the maximum mass of ammonia that can be produced from 280 kg of nitrogen. (2 marks)

Answer:

From equation: 1 mol N₂ produces 2 mol NH₃
Moles of NH₃ = 2 × 10,000 = 20,000 mol [1 mark]
Mass of NH₃ = 20,000 × 17 = 340,000 g = 340 kg [1 mark]

Marking notes: Award 1 mark for mole ratio application, 1 mark for final mass with units. Accept 340 kg or 340,000 g.

19. Titration to determine mass of potassium hydroxide.

(a) Calculate the number of moles of HCl used in the titration. (1 mark)

Answer:

Moles of HCl = 0.20 × (15.0 / 1000) = 0.20 × 0.0150 = 0.0030 mol [1 mark]

Marking notes: Award 1 mark for correct answer. Accept 0.003 mol.

(b) Calculate the concentration of the potassium hydroxide solution in mol/dm³. (2 marks)

Answer:

From equation: KOH + HCl → KCl + H₂O, mole ratio is 1:1
Moles of KOH in 25.0 cm³ = 0.0030 mol [1 mark]
Concentration of KOH = 0.0030 / (25.0 / 1000) = 0.0030 / 0.0250 = 0.12 mol/dm³ [1 mark]

Marking notes: Award 1 mark for recognising 1:1 mole ratio, 1 mark for correct concentration with units.

(c) Calculate the mass of potassium hydroxide the student originally dissolved in 100 cm³ of water. (2 marks)

Answer:

Moles of KOH in 100 cm³ = 0.12 × (100 / 1000) = 0.012 mol [1 mark]
Mᵣ of KOH = 39 + 16 + 1 = 56
Mass of KOH = 0.012 × 56 = 0.672 g = 0.67 g [1 mark]

Marking notes: Award 1 mark for moles in original solution, 1 mark for mass with units. Accept 0.67 g or 0.672 g.

20. Industrial production of ethanol.

(a) Calculate the relative molecular mass of ethanol. (1 mark)

Answer:

Mᵣ of C₂H₅OH = (2 × 12) + (6 × 1) + 16 = 24 + 6 + 16 = 46 [1 mark]

Marking notes: Award 1 mark for correct answer.

(b)(i) Calculate the theoretical yield of ethanol from 14.0 tonnes of ethene. (2 marks)

Answer:

Mᵣ of C₂H₄ = (2 × 12) + (4 × 1) = 24 + 4 = 28
Moles of C₂H₄ = (14.0 × 10⁶) / 28 = 500,000 mol = 5.0 × 10⁵ mol [1 mark]
From equation: 1 mol C₂H₄ produces 1 mol C₂H₅OH
Moles of C₂H₅OH = 500,000 mol
Mass of C₂H₅OH = 500,000 × 46 = 23,000,000 g = 23.0 tonnes [1 mark]

Marking notes: Award 1 mark for moles of ethene, 1 mark for theoretical mass with units. Accept 23 tonnes or 2.3 × 10⁷ g.

(b)(ii) Calculate the percentage yield of ethanol in this process. (2 marks)

Answer:

Percentage yield = (actual yield / theoretical yield) × 100 [1 mark]
Percentage yield = (18.4 / 23.0) × 100 = 80.0% [1 mark]

Marking notes: Award 1 mark for correct formula, 1 mark for correct answer. Accept 80%.

END OF ANSWER KEY