Edexcel A-Level Chemistry Paper 2, November 2021: Question 2

1 mark · Medium difficulty · Multiple Choice

Calculate the minimum volume of oxygen at room temperature and pressure needed for the complete combustion of 0.200 mol of butane given its balanced equation.

Practise this question

Question

Multiple choice question 2 showing the balanced equation for the complete combustion of butane: 2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(l). The question asks for the minimum volume of oxygen at r.t.p. needed for complete combustion of 0.200 mol of butane, given the molar volume of a gas at r.t.p. is 24.0 dm3 mol-1. Four multiple choice options are listed: A, 4.8 dm3; B, 9.6 dm3; C, 31.2 dm3; D, 62.4 dm3.
Question text

2 The equation for the complete combustion of butane is

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)

What is the minimum volume of oxygen, at room temperature and pressure (r.t.p.),

needed for the complete combustion of 0.200 mol of butane?

[Molar volume of a gas at r.t.p. = 24.0 dm3 mol−1]

A 4.8 dm3

B 9.6 dm3

C 31.2 dm3

D 62.4 dm3

(Total for Question 2 = 1 mark)

Mark scheme

Show the mark scheme Mark scheme for question 2 showing that the only correct answer is C (31.2 dm3), with explanations for why options A, B, and D are incorrect based on incorrect stoichiometric ratios.

Question

Answer Mark

Number

2 The only correct answer is C (31.2 dm3) (1)

A is not correct because the answer assumes a 1:1 ratio of butane to oxygen

B is not correct because the answer assumes a 1:2 ratio of butane to oxygen

D is not correct because the answer assumes a 1:13 ratio of butane to oxygen

How to answer it

Calculating Oxygen Volume in Alkane Combustion

What this question tests

This question assesses your ability to use balanced chemical equations to determine reacting mole ratios, apply molar gas volumes at room temperature and pressure (r.t.p.), and convert between moles and volumes of gases.

Question 2 (1 Mark)

The Complete Combustion of Butane

✅ Correct Answer

C (31.2 dm³)

Option C is the only correct answer because applying the stoichiometric ratio from the balanced equation yields the correct volume of oxygen gas.

💡 Key Knowledge

  • Molar Volume: At r.t.p., 1 mole of any gas occupies 24.0 dm³ .
  • Stoichiometry: Coefficients in balanced equations give direct mole ratios (e.g., 2 moles of C₄H₁₀ react with 13 moles of O₂ ).

🧠 Exam Technique

Always double-check the balancing numbers in the provided equation before performing ratio calculations. Do not assume a simple 1:1 or 1:2 ratio just because numbers look convenient.

📐 Step-by-Step Calculation

  1. Find moles of oxygen required:
    From the equation, 2 mol C₄H₁₀ react with 13 mol O₂ .
    Moles of O₂ = 0.200 × (13 / 2) = 1.30 mol .
  2. Convert moles of oxygen to volume:
    Volume = Moles × Molar Volume
    Volume = 1.30 mol × 24.0 dm³ mol⁻¹ = 31.2 dm³ .

❌ Common Errors & Distractor Analysis

  • Option A ( 4.8 dm³ ): Results from incorrectly assuming a 1:1 reacting ratio ( 0.200 × 24.0 ), ignoring the balancing coefficients.
  • Option B ( 9.6 dm³ ): Results from incorrectly assuming a 1:2 reacting ratio ( 0.200 × 2 × 24.0 ).
  • Option D ( 62.4 dm³ ): Results from using the oxygen stoichiometric coefficient directly as the mole value without dividing by the butane coefficient ( 0.200 × 13 × 24.0 ).
Examiner Note: Multiple-choice questions involving stoichiometry frequently use common stoichiometric miscalculations as incorrect distractors. Writing out the ratio clearly prevents these errors.

Topics

Physical Chemistry · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.