AQA A-Level Chemistry AS Paper 1, November 2021: Question 6

5 marks · Medium difficulty · Practical Techniques & Data Analysis

Calculate the mass of sulfur dioxide formed in the reaction between calcium sulfide and calcium sulfate, and show that calcium sulfate is the limiting reagent.

Practise this question

Question

Question 6 states that calcium sulfide reacts with calcium sulfate according to the equation: CaS + 3 CaSO4 -> 4 CaO + 4 SO2. It gives the masses of 2.50 g of calcium sulfide and 9.85 g of calcium sulfate heated together, and asks to show that calcium sulfate is the limiting reagent and to calculate the mass of sulfur dioxide formed. Relative molecular masses are provided: Mr(CaS) = 72.2 and Mr(CaSO4) = 136.2.
Question text

06 Calcium sulfide reacts with calcium sulfate as shown.

CaS + 3 CaSO4 → 4 CaO + 4 SO2

2.50 g of calcium sulfide are heated with 9.85 g of calcium sulfate until there is no

further reaction.

Show that calcium sulfate is the limiting reagent in this reaction.

Calculate the mass, in g, of sulfur dioxide formed.

Mr (CaS) = 72.2

Mr (CaSO4) = 136.2

[5 marks]

Mass of sulfur dioxide g

Mark scheme

Show the mark scheme Mark scheme for question 6 showing 5 marking points: M1 for calculating amount of CaS (0.0346 mol), M2 for amount of CaSO4 (0.0723 mol), M3 for justifying the limiting reagent by comparing the mole ratio, M4 for calculating moles of SO2 using stoichiometry (0.0964 mol), and M5 for calculating the final mass of SO2 (6.18 g).

Question Marking guidance Additional Comments/Guidelines Mark

2.50 M1: amount of CaS

amount of CaS = = 0.0346 mol 1

72.2

9.85

amount of CaSO4 = = 0.0723 mol M2: amount of CaSO 1

136.2 4

3 mol of CaSO4 needed for each mol of CaS, and M3: limiting reagent justification

n(CaSO4) is not 3 × n(CaO) 1

(so CaSO4 is the limiting reagent)

4 M4: moles of CaSO4 × 4/3 1

n(SO2) = n(CaSO4) × = 0.0964 mol

M5: M4 × 64.1 1

mass of SO2 = n(SO2) × 64.1 = 6.18g

If CaS used as limiting reagent then allow M4 and

M5 ecf.

Must look for M1 and M3

How to answer it

Limiting Reagents & Sulfur Dioxide Mass Calculation

🔍 What this question tests

This 5-mark quantitative chemistry question assesses your ability to calculate molar amounts from masses, deduce limiting reagents using stoichiometric mole ratios, use stoichiometric ratios to find product moles, and convert moles to mass with appropriate significant figures.

Question 06: Step-by-Step Solution & Examiner Guidance

CaS + 3CaSO₄ → 4CaO + 4SO₂ (5 marks total)

💡 Key Knowledge

  • Moles formula: Moles = Mass ÷ Mᵣ
  • Limiting Reagent: The reactant that is completely consumed, limiting the amount of product formed.
  • Stoichiometric Ratio: Compare actual reacting mole ratios to the balanced equation coefficients.

🧠 Exam Technique

  • Always calculate the moles of both reactants first when given two starting masses. Never guess which one is limiting!
  • Show clear comparisons (e.g., stating how many moles of one reactant are needed for the other).
  • Carry unrounded values through your intermediate calculation steps.

📐 Step-by-Step Calculation

Step 1 (M1): Calculate moles of calcium sulfide (CaS)

n(CaS) = 2.50 ÷ 72.2 = 0.0346 mol

Step 2 (M2): Calculate moles of calcium sulfate (CaSO₄)

n(CaSO₄) = 9.85 ÷ 136.2 = 0.0723 mol

Step 3 (M3): Justify the limiting reagent

From the balanced equation, 1 mol of CaS reacts with 3 mol of CaSO₄ .

If all the CaS reacted, we would need: 0.0346 × 3 = 0.104 mol of CaSO₄ .

However, we only have 0.0723 mol of CaSO₄ available. Therefore, CaSO₄ is the limiting reagent.

Step 4 (M4): Calculate moles of sulfur dioxide (SO₂) formed

Using the molar ratio between CaSO₄ and SO₂ (3 : 4):

n(SO₂) = n(CaSO₄) × (4 ÷ 3) = 0.0723 × (4 ÷ 3) = 0.0964 mol

Step 5 (M5): Calculate the mass of sulfur dioxide

Using Mᵣ of SO₂ = 64.1 :

Mass = 0.0964 × 64.1 = 6.18 g

Mark Breakdown: M1: Amount of CaS | M2: Amount of CaSO₄ | M3: Limiting reagent justification comparing ratios | M4: Correct mole scaling to SO₂ | M5: Final mass (Accept 6.18g - 6.2g depending on rounding intermediate values; ECF allowed if wrong limiting reagent was chosen with valid follow-through).

❌ Common Errors & Pitfalls

  • Comparing raw masses directly: Comparing 2.50 g and 9.85 g without converting to moles ignores the different molar masses ( Mᵣ ).
  • Inverting the ratio: Multiplying by 3 ÷ 4 instead of 4 ÷ 3 when finding the moles of SO₂ .
  • Premature rounding: Rounding intermediate mole values too aggressively, leading to final boundary value errors.

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.