Chemistry • Year 11 • Module 2 • Lesson 19
Module 2 Synthesis & Exam Practice
Apply stoichiometry, concentration and purity across multi-step problems; interpret real data and predict outcomes in new scenarios.
1. Sequence the calculation steps
A student needs to find the volume of H2 gas (at RTP) produced when a 95.0% pure sample of Mg is reacted with excess HCl. Balanced equation: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g). The eight steps below are shuffled. Write the correct order in the “Order” column. 8 marks (1 each)
| Order | Step description |
|---|---|
| State the final answer with correct units: V(H2) = ___ L at RTP. | |
| Apply percentage purity: m(pure Mg) = m(sample) × 0.950. | |
| Identify the mole ratio Mg : H2 = 1 : 1 from the balanced equation. | |
| Note the given mass of the Mg sample. | |
| Calculate n(H2) = n(Mg) × 1 (ratio 1:1). | |
| Write and balance the equation; identify the reactant and product of interest. | |
| Calculate n(Mg) = m(pure Mg) ÷ MM(Mg); MM(Mg) = 24.31 g mol−1. | |
| Convert n(H2) to volume: V = n × 24.8 L mol−1. |
2. Interpret titration data
A student titrated 25.0 mL aliquots of a NaOH solution of unknown concentration against 0.1000 mol L−1 HCl. The table below shows three trial results. Equation: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l). 9 marks
| Trial | Initial burette reading (mL) | Final burette reading (mL) | Volume HCl used (mL) |
|---|---|---|---|
| Rough | 0.00 | 24.40 | |
| 1 | 0.00 | 23.65 | |
| 2 | 23.65 | 47.35 | |
| 3 | 0.00 | 23.70 |
2.1 Complete the “Volume HCl used” column. 1 mark
2.2 Identify the concordant titres and calculate the average concordant titre. State which trial is excluded and why. 3 marks
2.3 Using the average titre, calculate n(HCl) used per titration. Show working. 2 marks
2.4 Hence calculate the concentration of the NaOH solution. Give your answer to 4 significant figures. 2 marks
2.5 The student notices the burette had a small air bubble trapped below the tap during Trial 1. Predict whether the recorded volume of HCl would be higher or lower than the true value, and classify this as a random or systematic error. 1 mark
3. Interpret a concentration–time graph
In a gravimetric analysis experiment, BaCl2(aq) was added to a 100 mL Na2SO4(aq) solution. The graph below shows the mass of BaSO4 precipitate collected versus the volume of 0.500 mol L−1 BaCl2(aq) added. Equation: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq). 7 marks
Figure 3.1. Mass of BaSO4 precipitate vs volume of BaCl2(aq) added (illustrative data).
3.1 Describe the trend shown in the graph. 2 marks
3.2 From the graph, read off the volume of BaCl2 added at the equivalence point. Calculate the moles of BaCl2 used (c = 0.500 mol L−1). 2 marks
3.3 Using the mole ratio and the plateau mass of BaSO4 (4.67 g), calculate the concentration of the original Na2SO4(aq) solution (V = 100 mL). MM(BaSO4) = 233.4 g mol−1. 3 marks
4. Predict and justify
A chemistry student dissolves 4.24 g of Na2CO3 (MM = 105.99 g mol−1) in water and makes it up to 200 mL. She then adds this solution to excess CaCl2(aq). Equation: Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2NaCl(aq). 5 marks
4.1 Calculate the mass of CaCO3 precipitate expected (theoretical yield). MM(CaCO3) = 100.09 g mol−1. Show all steps. 3 marks
4.2 The student collects 3.82 g of dry CaCO3. Predict whether the percentage yield will be above or below 100% and calculate it. Suggest one reason the yield is not 100%. 2 marks
Q1 — Sequence the steps (correct order)
- Note the given mass of the Mg sample.
- Write and balance the equation; identify the reactant and product of interest.
- Apply percentage purity: m(pure Mg) = m(sample) × 0.950.
- Calculate n(Mg) = m(pure Mg) ÷ MM(Mg); MM(Mg) = 24.31 g mol−1.
- Identify the mole ratio Mg : H2 = 1 : 1 from the balanced equation.
- Calculate n(H2) = n(Mg) × 1 (ratio 1:1).
- Convert n(H2) to volume: V = n × 24.8 L mol−1.
- State the final answer with correct units: V(H2) = ___ L at RTP.
Q2.1 — Volume column
Rough: 24.40 mL; Trial 1: 23.65 mL; Trial 2: 47.35 − 23.65 = 23.70 mL; Trial 3: 23.70 mL.
Q2.2 — Concordant titres and average
Concordant titres: Trials 1, 2 and 3 (23.65, 23.70, 23.70 mL — all within 0.10 mL of each other). The Rough titre (24.40 mL) is excluded because it is used only to approximate the endpoint; it is not performed with care and differs by >0.10 mL. Average concordant titre = (23.65 + 23.70 + 23.70) ÷ 3 = 23.68 mL.
Q2.3 — n(HCl)
V(HCl) = 23.68 mL = 0.02368 L; n(HCl) = c × V = 0.1000 × 0.02368 = 2.368 × 10−3 mol.
Q2.4 — c(NaOH)
Ratio NaOH : HCl = 1 : 1; n(NaOH) = 2.368 × 10−3 mol per 25.0 mL aliquot. c(NaOH) = n ÷ V = 2.368 × 10−3 ÷ 0.02500 = 0.09472 mol L−1.
Q2.5 — Air bubble error
An air bubble that disappears during the titration means the student is also dispensing that air volume. The recorded HCl volume would be higher than the true volume used. This would affect only the trial where the bubble was present, so it is a random error (unless the student always traps a bubble, in which case it becomes systematic).
Q3.1 — Trend
The mass of BaSO4 precipitate increases linearly from 0 g to approximately 4.67 g as the volume of BaCl2 added increases from 0 to 40 mL. Beyond 40 mL the mass remains constant (plateau), indicating all Na2SO4 has been consumed.
Q3.2 — Equivalence point moles
Equivalence point ≈ 40 mL = 0.0400 L. n(BaCl2) = 0.500 × 0.0400 = 0.0200 mol.
Q3.3 — c(Na2SO4)
n(BaSO4) = 4.67 ÷ 233.4 = 0.02001 mol. Ratio BaCl2 : Na2SO4 = 1 : 1, so n(Na2SO4) = 0.02001 mol. V = 100 mL = 0.100 L. c(Na2SO4) = 0.02001 ÷ 0.100 = 0.200 mol L−1.
Q4.1 — Theoretical yield CaCO3
n(Na2CO3) = 4.24 ÷ 105.99 = 0.04001 mol. Ratio Na2CO3 : CaCO3 = 1 : 1; n(CaCO3) = 0.04001 mol. m(CaCO3) = 0.04001 × 100.09 = 4.005 g.
Q4.2 — Percentage yield
% yield = (3.82 ÷ 4.005) × 100 = 95.4% — below 100% (as expected; yield can never exceed 100%). Possible reasons: not all precipitate was collected during filtration; some BaSO4 remained in solution (slight solubility); small amounts lost during washing or drying.
Marking criteria: 1 mark for correctly identifying % yield < 100% and calculating it; 1 mark for a valid, specific reason.