AQA A-Level Chemistry AS Paper 1, June 2025: Question 13

1 mark · Medium difficulty · Multiple Choice

Calculate the maximum mass of sodium chloride produced from given masses of sodium and chlorine by identifying the limiting reagent.

Practise this question

Question

Multiple-choice question 13 asks: 'What is the maximum mass of sodium chloride that can be made from 10.0 g of sodium and 10.0 g of chlorine?' Four options are given: A 8.24 g, B 16.5 g, C 20.0 g, and D 25.4 g.
Question text

13 What is the maximum mass of sodium chloride that can be made from

10.0 g of sodium and 10.0 g of chlorine?

[1 mark]

A 8.24 g

B 16.5 g

C 20.0 g

D 25.4 g

Mark scheme

Show the mark scheme Mark scheme table row for question 13 showing the correct answer as B, worth 1 mark (AO2), corresponding to 16.5 g.

13 B 1 (AO2) 16.5 g

How to answer it

Calculating Maximum Mass (Limiting Reactant)

📋 What this question tests

This question assesses your ability to apply stoichiometry and the concept of limiting reactants to find theoretical yield. Key skills include: writing a balanced equation for the reaction between sodium and chlorine, converting mass to moles, identifying the limiting reagent, and calculating the mass of the product formed using molar mass.

Question 13

Theoretical Yield of Sodium Chloride

Multiple Choice Question (1 Mark)

✅ Correct Answer

B — 16.5 g

Mark allocation: 1 mark (AO2). Requires correct identification of chlorine gas as diatomic (Cl₂), determination of the limiting reagent, and subsequent calculation of the theoretical yield of NaCl.

💡 Key Knowledge

  • Diatomic Element: Chlorine exists as Cl₂ with Mr = 2 × 35.5 = 71.0 g mol⁻¹ .
  • Balanced Chemical Equation:
    2Na + Cl₂ → 2NaCl
  • Limiting Reactant Principle: The theoretical yield is entirely determined by whichever reactant runs out first based on the mole ratio.

📐 Step-by-Step Calculation

  1. Calculate the moles of each reactant available:
    Moles of Na = mass / Ar = 10.0 / 23.0 = 0.435 mol
    Moles of Cl₂ = mass / Mr = 10.0 / 71.0 = 0.141 mol
  2. Determine the limiting reactant:
    From the balanced equation ( 2Na + Cl₂ → 2NaCl ), the required reacting ratio is 2 mol Na : 1 mol Cl₂ .
    Moles of Na needed to react with all Cl₂ = 2 × 0.141 mol = 0.282 mol .
    Since we have 0.435 mol Na (which is more than 0.282 mol ), sodium is in excess and chlorine (Cl₂) is the limiting reactant.
  3. Calculate the theoretical yield of NaCl:
    Mole ratio is 1 mol Cl₂ : 2 mol NaCl .
    Moles of NaCl formed = 2 × 0.1408 mol = 0.2817 mol .
  4. Convert moles of NaCl to mass:
    Mr(NaCl) = 23.0 + 35.5 = 58.5 g mol⁻¹
    Mass of NaCl = 0.2817 mol × 58.5 g mol⁻¹ = 16.48 g ≈ 16.5 g (3 s.f.)

❌ Common Errors & Distractor Analysis

  • Option A (8.24 g): Occurs if you forget the 1 : 2 mole ratio between Cl₂ and NaCl, calculating 0.1408 × 58.5 = 8.24 g .
  • Option C (20.0 g): Occurs if you naively apply the conservation of mass ( 10.0 g + 10.0 g ), assuming both reactants are completely consumed without checking stoichiometry.
  • Option D (25.4 g): Occurs if you incorrectly assume sodium is limiting ( 10.0 / 23.0 = 0.435 mol × 58.5 = 25.4 g ).

🧠 Exam Technique & Examiner Tips

  • Two Masses Given = Limiting Reagent Problem: Whenever you are given masses for both starting materials in a question, immediately calculate moles for both to find which one limits the yield.
  • Watch Diatomic Elements: "Chlorine" means the element in its standard state, Cl₂ ( Mr = 71.0 ), not isolated chlorine atoms ( 35.5 ).
  • Quick Sanity Check: With equal masses (10.0 g each), chlorine has a much higher molar mass per reactive unit ( 71.0 vs 2 × 23.0 = 46.0 ), so chlorine must run out first.

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.