OCR A-Level Chemistry AS Breadth in chemistry (01), June 2019: Question 5

1 mark · Medium difficulty · Multiple Choice

Calculate the total volume of the gas mixture at room temperature and pressure after the reaction between nitric oxide and oxygen is complete.

Practise this question

Question

Multiple-choice question 5 asks for the total volume of gas mixture at RTP after mixing 8.0 dm3 of NO with 6.0 dm3 of O2 according to the equation 2NO(g) + O2(g) -> 2NO2(g). Four multiple-choice options are provided: A (8.0 dm3), B (10.0 dm3), C (12.0 dm3), and D (14.0 dm3), followed by a box for the answer and a mark allocation of [1].
Question text

58.0 dm3 of NO is mixed with 6.0 dm3 of O at room temperature and pressure (RTP).

The reaction below takes place until one of the reactants is used up.

2NO(g) + O2(g) → 2NO2(g)

What is the volume of the mixture at RTP after the reaction has taken place?

A 8.0 dm3

B 10.0 dm3

C 12.0 dm3

D 14.0 dm3

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme table shows question number 5 with the correct answer B, worth 1 mark, under assessment objective AO2.6.

5 B 1 AO2.6

How to answer it

Gas Volumes & Limiting Reactants Study Guide

What this question tests

This question assesses your ability to apply Avogadro's Law (that gases of the same volume at the same temperature and pressure contain the same number of moles), identify limiting reagents in gaseous systems, calculate stoichiometric reaction changes, and determine the total residual volume of a gaseous mixture after a reaction has completed.

Question 5 Multiple Choice Analysis

Determining Total Gas Volume After Reaction

✅ Correct Answer: B (10.0 dm³)

Option B is correct because NO acts as the limiting reactant, leaving unreacted O₂ and producing new NO₂ gas in the container.

💡 Key Knowledge

  • At room temperature and pressure (RTP), gas volumes are directly proportional to moles (Avogadro's Law). You can use volumes directly in place of moles for stoichiometric ratios.
  • A limiting reactant is completely consumed and limits the amount of product formed.

🧠 Exam Technique

For gas mixture questions involving a reaction, always follow a systematic 3-step approach: (1) Identify the limiting reagent using the stoichiometric ratio, (2) calculate amounts of unreacted excess and products formed, and (3) sum up all gases present in the final vessel.

❌ Common Errors

Students often forget to add the remaining excess reactant back into the final total volume, or they incorrectly assume that all reactants are completely used up (forgetting excess reagents).

📐 Step-by-Step Calculation

  1. Examine the balanced equation: 2NO(g) + O₂(g) → 2NO₂ (g)
    Ratio is 2 : 1 : 2
  2. Identify the limiting reagent:
    We have 8.0 dm³ of NO and 6.0 dm³ of O₂ .
    By stoichiometry, 8.0 dm³ of NO requires 8.0 ÷ 2 = 4.0 dm³ of O₂ .
    Since we have 6.0 dm³ of O₂ available, O₂ is in excess, making NO the limiting reagent.
  3. Calculate remaining excess reactant:
    Volume of O₂ used = 4.0 dm³ .
    Volume of unreacted O₂ remaining = 6.0 dm³ - 4.0 dm³ = 2.0 dm³ .
  4. Calculate product formed:
    From the 2 : 2 ratio, 8.0 dm³ of NO produces 8.0 dm³ of NO₂ .
  5. Calculate total final volume:
    Total volume = (Unreacted O₂ ) + (Produced NO₂ )
    Total volume = 2.0 dm³ + 8.0 dm³ = 10.0 dm³ (Option B).
Mark awarded: 1 mark for identifying B (AO2.6: Applying chemical knowledge and quantitative skills to unfamiliar scenarios).

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), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.