OCR A-Level Chemistry AS Breadth in chemistry (01), November 2020: Question 8

1 mark · Medium difficulty · Multiple Choice

Determine the number of moles of HNO3 that react with 1 mole of Cu in the unbalanced equation for the reaction of copper with concentrated nitric acid.

Practise this question

Question

Multiple choice question 8 showing an unbalanced chemical equation for the reaction of copper with concentrated nitric acid: Cu + HNO3 -> Cu(NO3)2 + NO2 + H2O. Four options are provided: A 2, B 3, C 4, D 6, along with an answer box.
Question text

8 The unbalanced equation for the reaction of copper with concentrated nitric acid is shown below.

… Cu + … HNO3 → … Cu(NO3)2 + … NO2 + … H2O

What is the number of moles of HNO3 that react with 1 mole of Cu?

A 2

B 3

C 4

D 6

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme for question 8 showing the correct answer is C, corresponding to 4 moles.

8 C 1 1.2 ALLOW 4

How to answer it

Balancing Redox Equations: Copper & Nitric Acid

What this question tests

This question assesses your ability to balance a complex redox equation by determining stoichiometry, applying reacting ratios, and interpreting chemical formulae in the context of transition metals reacting with concentrated acids.

Question 8: Multiple Choice Solution

Reaction of Copper with Concentrated Nitric Acid

✅ Correct Answer

C (4)

When the equation is fully balanced, 1 mole of Cu reacts with 4 moles of HNO₃ .

💡 Key Knowledge

  • Concentrated nitric acid acts as an oxidizing agent, producing brown nitrogen dioxide gas ( NO₂ ).
  • Copper is oxidized from state 0 to +2 in Cu(NO₃)₂ .
  • Nitrogen is reduced from +5 to +4 in NO₂ .

🧠 Exam Technique

For complex equations that are tricky to balance by pure inspection, start by balancing the element that changes oxidation state or appears in the fewest compounds (like Cu ), then tackle nitrogen and oxygen systematically.

❌ Common Errors

Students frequently confuse the reaction of copper with dilute nitric acid (which produces nitric oxide, NO ) or miscount the total number of nitrate ions needed on the reactant side to account for both the salt and the reduction product.

📐 Step-by-Step Balancing Guide

  1. Start with Copper: Cu + ...HNO₃ → Cu(NO₃)₂ + ...NO₂ + ...H₂O (Already balanced with 1 Cu on each side).
  2. Count Nitrate Groups: The copper(II) nitrate product contains 2 nitrate ions ( NO₃⁻ ). Furthermore, each NO₂ molecule accounts for a reduced nitrogen. Balancing the redox transfer shows we need 2 molecules of NO₂ for every copper oxidized.
  3. Combine Reacting Proportions: Total NO₃⁻ needed = 2 (in salt) + 2 (as NO₂ gas) = 4 HNO₃ molecules.
  4. Balance Hydrogen & Oxygen: Place a coefficient of 2 in front of H₂O to balance 4 hydrogens:
    Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O .
Mark Allocation: [1 mark] awarded for selecting option C.

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.