OCR A-Level Chemistry AS Breadth in chemistry (01), June 2023: Question 8
1 mark · Medium difficulty · Multiple Choice
Calculate the total volume of gas produced at RTP when 0.00250 mol of magnesium nitrate is thermally decomposed.
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Question text
8 Magnesium nitrate, Mg(NO3)2, decomposes when heated:
Mg(NO3)2(s) MgO(s) + 2NO2(g) + ½O2(g)
0.00250 mol of Mg(NO3)2 is decomposed.
What is the volume of gas produced, measured at RTP?
A 30 cm3
B 60 cm3
C 120 cm3
D 150 cm3
Your answer [1]
Mark scheme
Show the mark scheme
8 D 1 AO2.6
How to answer it
Thermal Decomposition of Magnesium Nitrate
This question assesses your understanding of Group 2 thermal decomposition reactions, stoichiometric mole ratios in balanced chemical equations, and the conversion between moles of gas and volume at room temperature and pressure (RTP).
Question 8 (Multiple Choice)
Full Worked Solution & Examiner Breakdown
✅ Correct Answer
D (150 cm³)
💡 Key Knowledge
- Group 2 nitrates decompose on heating to form a metal oxide, nitrogen dioxide gas, and oxygen gas.
- One mole of any gas occupies 24.0 dm³ (or 24,000 cm³ ) at RTP.
- State symbols matter: check which products are gases ( g ) versus solids ( s ).
🧠 Exam Technique
Do not rush multiple-choice gas volume questions. Always sum the total moles of all gaseous products from the stoichiometric coefficients before multiplying by the molar gas volume.
❌ Common Errors
- Only counting NO₂: Forgetting to include the O₂ gas produced, leading to an incorrect total mole ratio.
- Unit conversion traps: Forgetting to multiply by 24,000 to convert dm³ into cm³.
- Ratio inversion: Misreading the 1 : (2 + 0.5) stoichiometric ratio.
📐 Step-by-Step Calculation
- Identify total moles of gas per mole of Mg(NO₃)₂:
From the equation Mg(NO₃)₂(s) → MgO(s) + 2NO₂(g) + 0.5O₂(g) , 1 mole of Mg(NO₃)₂ produces 2 + 0.5 = 2.5 moles of gas in total. - Calculate total moles of gas produced from 0.00250 mol of reactant:
Total moles of gas = 0.00250 mol × 2.5 = 0.00625 mol - Convert moles of gas to volume at RTP (in cm³):
Volume = Moles × Molar Gas Volume ( 24,000 cm³ mol⁻¹ )
Volume = 0.00625 mol × 24,000 cm³ mol⁻¹ = 150 cm³
Topics
Module 2: Foundations in chemistry · Module 3: Periodic table and energy · 2.1 Atoms and reactions · 3.1 The periodic table
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.