OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2022: Question 2

1 mark · Medium difficulty · Multiple Choice

Identify the correct equation for the reaction of 20 cm3 of nitrogen gas with 10 cm3 of oxygen gas to form 20 cm3 of a gaseous product.

Practise this question

Question

Multiple-choice question 2 states that 20 cm3 of nitrogen gas reacts with 10 cm3 of oxygen gas to form 20 cm3 of a gaseous product, and asks which equation is the most likely. Four options are given: A, N2(g) + O2(g) -> 2NO(g); B, N2(g) + 2O2(g) -> N2O4(g); C, 2N2(g) + O2(g) -> 2N2O(g); D, 2N2(g) + 2O2(g) -> 4NO(g). An answer box is provided at the bottom.
Question text

2 20 cm3 of nitrogen gas reacts with 10 cm3 of oxygen gas to form 20 cm3 of a gaseous product.

Which equation is the most likely for the reaction?

A N2(g) + O2(g) 2NO(g)

B N2(g) + 2O2(g) N2O4(g)

C 2N2(g) + O2(g) 2N2O(g)

D 2N2(g) + 2O2(g) 4NO(g)

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is C, worth 1 mark.

2 C 1 1.2

How to answer it

Reacting Volumes of Gases

What this question tests

This question assesses your understanding of Avogadro's Law, which states that under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules. You are required to translate reacting gas volumes directly into a simplified molar ratio to determine the correct balanced chemical equation.

Question Multiple Choice Analysis

Question 2

✅ Correct Answer

C

Equation: 2N₂(g) + O₂(g) → 2N₂O(g)

💡 Key Knowledge

  • Avogadro's Principle: Gas volume is directly proportional to moles at constant temperature and pressure.
  • Volume ratio = Molar ratio (stoichiometric coefficients).
  • Simplifying experimental gas volumes gives the simplest whole-number equation coefficients.

📐 Step-by-Step Calculation

  1. Identify given volumes:
    Volume of N₂ = 20 cm³
    Volume of O₂ = 10 cm³
    Volume of product = 20 cm³
  2. Write initial ratio (N₂ : O₂ : Product):
    20 : 10 : 20
  3. Simplify to lowest whole numbers:
    Divide through by 10:
    2 : 1 : 2
  4. Match coefficients to options: Look for an equation with stoichiometric coefficients of 2 for N₂, 1 for O₂, and 2 for the gaseous product, which points directly to option C.

🧠 Exam Technique

Don't panic about complex stoichiometry or molar gas volumes (like 24.0 dm³ mol⁻¹). Because all reactants and products are gases measured under the same conditions, you can completely bypass calculating actual moles. Just convert the given volumes straight into a ratio!

❌ Common Errors

  • Reversing ratios: Mixing up reactants and products or assigning the 10 cm³ volume to nitrogen instead of oxygen.
  • Ignoring simplification: Forgetting to reduce the volume ratio 20 : 10 : 20 down to 2 : 1 : 2 , leading confusion among options A, C, and D.
Mark Scheme 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, Periodic table, elements and physical chemistry (01), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.