AQA A-Level Chemistry Paper 3, 2017: Question 19

1 mark · Medium difficulty · Multiple Choice

Calculate the minimum mass in mg of zinc needed to react completely with 50.0 cm³ of 1.68 mol dm⁻³ hydrochloric acid.

Practise this question

Question

Question 19 presents the balanced equation Zn + 2HCl -> ZnCl2 + H2. It asks for the minimum mass, in mg, of zinc (Ar = 65.4) needed to react with 50.0 cm³ of 1.68 mol dm⁻³ hydrochloric acid. Four multiple-choice options are given: A is 2.75, B is 5.49, C is 2.75 x 10³, and D is 5.49 x 10³.
Question text

19 The equation for the reaction between zinc and hydrochloric acid is

Zn + 2HCl → ZnCl2 + H2

What is the minimum mass, in mg, of zinc (Ar = 65.4) needed to react with

50.0 cm3 of 1.68 mol dm−3 hydrochloric acid?

[1 mark]

A 2.75

B 5.49

C 2.75 × 103

D 5.49 × 103

Mark scheme

Show the mark scheme Mark scheme table showing question number 19 with correct answer C.

19 C

How to answer it

Calculating Mass from Solution Concentration & Stoichiometry

📋 What this question tests

This question assesses your core quantitative chemistry skills from Physical Chemistry (Amount of Substance):

  • Calculating moles in a solution using n = c × V .
  • Applying reacting stoichiometric mole ratios from a balanced chemical equation.
  • Converting moles of a solid element to mass in grams using m = n × Aᵣ .
  • Carrying out metric unit conversions ( cm³ → dm³ and g → mg ).

Question 19

Multiple Choice (1 Mark)

✅ Correct Option: C (2.75 × 10³)

The minimum mass of zinc required is 2.75 × 10³ mg (or 2750 mg, which equals 2.75 g).

📐 Step-by-Step Calculation

Reaction Equation: Zn + 2HCl → ZnCl₂ + H₂

1 Calculate moles of HCl reacting:
Volume in dm³ = 50.0 / 1000 = 0.0500 dm³
n(HCl) = c × V = 1.68 mol dm⁻³ × 0.0500 dm³ = 0.0840 mol

2 Use the mole ratio to find moles of Zn:
From the equation, Zn : HCl = 1 : 2
n(Zn) = n(HCl) ÷ 2 = 0.0840 ÷ 2 = 0.0420 mol

3 Calculate mass of Zn in grams (g):
mass (g) = n × Aᵣ = 0.0420 mol × 65.4 g mol⁻¹ = 2.7468 g

4 Convert mass to milligrams (mg) and round to 3 s.f.:
mass (mg) = 2.7468 g × 1000 = 2746.8 mg = 2.75 × 10³ mg

💡 Key Knowledge

  • Concentration formula: n = (c × V in cm³) / 1000
  • Mass formula: m = n × Mᵣ (or Aᵣ for atoms)
  • Unit conversions:
    • 1 dm³ = 1000 cm³
    • 1 g = 1000 mg = 10³ mg
  • Ratios: Always look at the balancing number in front of each species in the equation.

❌ Common Errors & Distractor Breakdown

  • Option A (2.75): Calculated the mass correctly in grams (2.75 g) but failed to convert to mg.
  • Option B (5.49): Failed to divide by 2 for the 1:2 ratio and also forgot to convert g to mg ( 0.0840 × 65.4 = 5.49 g ).
  • Option D (5.49 × 10³): Successfully converted to mg, but forgot the 1 : 2 reacting ratio ( 0.0840 × 65.4 × 1000 = 5.49 × 10³ mg ).

🧠 Exam Technique & Multiple Choice Strategy

  • Underline units immediately: The question specifically asked for mg , not g . Examiners deliberately pair distractors in pairs differing by a factor of 10³ ( 2.75 vs 2.75 × 10³ ).
  • Spot the pairing pattern: Notice how A and C share the coefficient 2.75 , while B and D share 5.49 ( 2 × 2.75 ). This tells you immediately that the two hurdles are (1) remembering the 1:2 ratio, and (2) applying the ×10³ factor for milligrams.
  • Sanity check: 50 cm³ of ~1.7 mol dm⁻³ acid contains under 0.1 mol of solute, which should react with roughly 0.04 mol of metal. 0.04 mol × ~65 g mol⁻¹ is roughly 2.6 g = 2600 mg. An order of magnitude check rules out A and B instantly!

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.