OCR A-Level Chemistry AS Breadth in chemistry (01), June 2023: Question 9

Calculating Mass from Limiting Reagents

OCR AS Level Chemistry • Multiple Choice Question

What this question tests

This question assesses your ability to calculate moles from mass and solution concentration, identify the limiting reagent in a chemical reaction using stoichiometric ratios, and use the reacting mole ratio to determine the mass of a product formed.

Question Analysis & Answer

Question 9

✅ Correct Answer: C (0.33 g)

By determining the moles of both reactants and comparing them using the stoichiometric equation, you find that silver nitrate is the limiting reagent, which yields 0.003075 moles of silver metal (approx. 0.33 g).

💡 Key Knowledge

Formula: Moles = Concentration × Volume (in dm³)
Formula: Moles = Mass ÷ Molar Mass (A_r)
Stoichiometry: Always check the balanced equation coefficients before identifying the limiting reagent.

🧠 Exam Technique

In quantitative multiple-choice questions, never guess! Write down your working space methodically. Always convert volumes from cm³ to dm³ by dividing by 1000 early in your calculation to avoid scale errors.

❌ Common Errors

Ignoring stoichiometry: Forgetting to divide moles of AgNO₃ by 2 when comparing it to Zn.
Unit blindness: Forgetting to convert 15 cm³ into dm³ (using 15 instead of 0.015).
Wrong limiting reagent: Assuming the reactant with the smaller mass or smaller initial moles is automatically the limiting reagent.

Step-by-Step Calculation

Working out why C is correct

📐 Step 1: Calculate moles of Zinc (Zn) added

Mass of Zn = 0.10 g

Molar mass (A_r) of Zn = 65.4 g mol⁻¹

Moles of Zn = 0.10 ÷ 65.4 = 0.001529 mol

📐 Step 2: Calculate moles of Silver Nitrate (AgNO₃) available

Volume = 15 cm³ = 0.015 dm³

Concentration = 0.25 mol dm⁻³

Moles of AgNO₃ = 0.25 × 0.015 = 0.00375 mol

📐 Step 3: Identify the Limiting Reagent

From the balanced equation: Zn(s) + 2AgNO₃(aq) → 2Ag(s) + Zn(NO₃)₂

The reaction requires 2 moles of AgNO₃ for every 1 mole of Zn.

Required AgNO₃ for all Zn to react = 0.001529 mol × 2 = 0.003058 mol.

Since we have 0.00375 mol of AgNO₃ available (which is more than 0.003058 mol), AgNO₃ is in excess, making Zinc (Zn) the limiting reagent.

Alternative check: Reacting ratio of AgNO₃ to Zn is 2:1. Available moles: Zn = 0.001529, AgNO₃ = 0.00375. Dividing by coefficients: Zn = 0.001529, AgNO₃ = 0.001875. Since 0.001529 is smaller, Zn limits the reaction.

📐 Step 4: Calculate the mass of Silver (Ag) produced

From the balanced equation, 1 mole of Zn produces 2 moles of Ag.

Moles of Ag produced = Moles of Zn reacted × 2

Moles of Ag = 0.001529 × 2 = 0.003058 mol

Molar mass (A_r) of Ag = 107.9 g mol⁻¹

Mass of Ag = Moles × Molar Mass = 0.003058 × 107.9 = 0.3299 g

Rounding to 2 significant figures (matching input data like 0.10 g and 15 cm³) gives 0.33 g.

Examiner Note: Top-level responses quickly evaluated the reacting ratios rather than blindly calculating product masses for both reagents. Option C successfully traps students who correctly navigate the limiting reagent steps.