AQA A-Level Chemistry Paper 3, 2019: Question 7

1 mark · Medium difficulty · Multiple Choice

Determine the amount in moles of lead(II) iodide formed when given volumes and concentrations of lead(II) nitrate and potassium iodide are reacted.

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Question

Question 07 shows the balanced chemical equation Pb(NO3)2(aq) + 2KI(aq) -> PbI2(s) + 2KNO3(aq). It states that 25.0 cm³ of a 0.100 mol dm⁻³ solution of each compound are mixed together, and asks which amount, in mol, of lead(II) iodide is formed. Four multiple-choice options are given: A (1.25 x 10⁻³), B (2.50 x 10⁻³), C (1.25 x 10⁻²), and D (2.50 x 10⁻²).
Question text

07 Lead(II) nitrate and potassium iodide react according to the equation

Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)

In an experiment, 25.0 cm3 of a 0.100 mol dm–3 solution of each compound are

mixed together.

Which amount, in mol, of lead(II) iodide is formed?

[1 mark]

A 1.25 x 10–3

B 2.50 x 10–3

C 1.25 x 10–2

D 2.50 x 10–2

Mark scheme

Show the mark scheme Mark scheme table row for question 7, indicating the correct answer is option A with 1 mark awarded.

7 A 1

How to answer it

Amount of Substance: Limiting Reactant Calculation

What this question tests

This multiple-choice question tests your ability to calculate moles in solution ( n = c × V ), identify the limiting reactant using stoichiometric mole ratios from a balanced equation, and deduce the maximum theoretical yield of a solid precipitate in moles.

Question 07 • 1 Mark

Lead(II) Iodide Precipitation

Reaction equation: Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

✅ Correct Answer: A

1.25 × 10⁻³ mol

Mark Scheme: Option A = 1 Mark (No partial credit).

💡 Key Knowledge

  • Solution formula: moles = concentration (mol dm⁻³) × volume (dm³)
  • Volume conversion: 1 dm³ = 1000 cm³, so divide cm³ by 1000.
  • Stoichiometric Ratio: 1 mole of Pb(NO₃)₂ reacts with 2 moles of KI to produce 1 mole of PbI₂.
  • Limiting Reactant Rule: The reactant completely consumed first dictates the maximum amount of product that can form.

📐 Step-by-Step Calculation

Step 1: Calculate moles of each reactant added
Both solutions have the same volume (25.0 cm³) and concentration (0.100 mol dm⁻³):
Volume = 25.0 / 1000 = 0.0250 dm³
n(Pb(NO₃)₂) = 0.100 mol dm⁻³ × 0.0250 dm³ = 2.50 × 10⁻³ mol
n(KI) = 0.100 mol dm⁻³ × 0.0250 dm³ = 2.50 × 10⁻³ mol
Step 2: Identify the limiting reactant
The stoichiometric ratio is 1 Pb(NO₃)₂ : 2 KI .
To react completely with 2.50 × 10⁻³ mol of Pb(NO₃)₂, you would require:
2 × (2.50 × 10⁻³) = 5.00 × 10⁻³ mol of KI .
However, you only have 2.50 × 10⁻³ mol of KI . Therefore, KI is the limiting reagent (and Pb(NO₃)₂ is in excess).
Step 3: Calculate the moles of PbI₂ formed
Look at the molar ratio between the limiting reactant (KI) and the product (PbI₂):
2 mol KI → 1 mol PbI₂ (a 2 : 1 ratio)
n(PbI₂) = n(KI) / 2 = (2.50 × 10⁻³) / 2 = 1.25 × 10⁻³ mol

❌ Common Distractor Traps

  • Choosing B (2.50 × 10⁻³ mol): The most common error! Students notice a 1:1 ratio between Pb(NO₃)₂ and PbI₂ and assume all 2.50 × 10⁻³ mol reacts, forgetting to check whether there is enough KI present.
  • Choosing C (1.25 × 10⁻² mol) or D (2.50 × 10⁻² mol): A standard power-of-ten unit conversion blunder caused by dividing by 100 instead of 1000 when converting cm³ to dm³.

🧠 Exam Technique & Examiner Tips

  • Two amounts given? Always check limiting reagent! Whenever an exam question specifies volumes and concentrations for both reactants, it is almost certainly a limiting reactant question.
  • Divide moles by stoichiometric coefficient: A rapid shortcut to find the limiting reagent:
    • Pb(NO₃)₂: 2.50 × 10⁻³ / 1 = 2.50 × 10⁻³
    • KI: 2.50 × 10⁻³ / 2 = 1.25 × 10⁻³ (smallest value = limiting reagent!)
  • The smallest calculated value directly gives the theoretical yield of any product with a stoichiometric coefficient of 1 (like PbI₂).

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.