Chemistry • Year 11 • Module 2 • Lesson 16
Stoichiometry in Solution
Build HSC Band 5–6 extended-response technique on multi-step solution stoichiometry, excess-reagent analysis, and evaluation of experimental errors.
1. Multi-step calculation — copper sulfate reaction with sodium hydroxide (Band 4–6)
8 marks Band 4–6
Scenario. A chemistry student mixes 40.0 mL of 0.150 mol/L CuSO4 solution with 60.0 mL of 0.150 mol/L NaOH solution. The balanced equation is:
CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq)
Molar masses: Cu = 63.546, O = 15.999, H = 1.008, Na = 22.990, S = 32.06 g/mol.
Q1. Answer all four parts, showing all steps and unit conversions.
(a) Calculate the moles of CuSO4 and NaOH. (2 marks)
(b) Determine which reagent is limiting. Show your full comparison using the n ÷ coefficient method. (2 marks)
(c) Calculate the mass of Cu(OH)2 precipitate formed. (2 marks)
(d) Calculate the concentration of Na2SO4 in the final 100.0 mL solution. Explain why you use 100.0 mL rather than 40.0 mL or 60.0 mL. (2 marks)
2. Data + scenario: evaluating a student’s solution-stoichiometry working (Band 5–6)
8 marks Band 5–6
Scenario. A student is asked to find the concentration of an unknown H2SO4 solution. They use 25.0 mL of the unknown H2SO4 solution and react it with 40.0 mL of 0.100 mol/L NaOH. The balanced equation is: H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l). The student’s working is shown below.
Student’s working:
Step 1: n(NaOH) = 0.100 × 40.0 = 4.00 mol
Step 2: Ratio 1:1 → n(H2SO4) = 4.00 mol
Step 3: c(H2SO4) = 4.00 ÷ 0.0250 = 160 mol/L
Q2. Analyse and evaluate the student’s working. In your response you must:
- Identify every error in the student’s three steps, explaining what is wrong in each case.
- Provide the correct working showing all steps, including all unit conversions.
- State the correct concentration of the unknown H2SO4 solution.
- Explain how the student could use the “reasonableness check” to detect that their answer of 160 mol/L is implausible.
- Identify one experimental precaution that improves the reliability of this type of titration result.
Q1 — Model answer and marking criteria (8 marks)
(a) Moles of each reactant (2 marks):
n(CuSO4) = 0.150 × 0.0400 = 0.00600 mol [1].
n(NaOH) = 0.150 × 0.0600 = 0.00900 mol [1].
(b) Limiting reagent comparison (2 marks):
CuSO4: 0.00600 ÷ 1 = 0.00600. NaOH: 0.00900 ÷ 2 = 0.00450 [1].
NaOH gives the smaller value (0.00450 < 0.00600), so NaOH is the limiting reagent. CuSO4 is in excess [1].
(c) Mass of Cu(OH)2 (2 marks):
Ratio NaOH : Cu(OH)2 = 2:1; n(Cu(OH)2) = 0.00900 ÷ 2 = 0.00450 mol [1].
MM(Cu(OH)2) = 63.546 + 2(15.999 + 1.008) = 97.560 g/mol; m = 0.00450 × 97.560 = 0.439 g [1].
(d) Concentration of Na2SO4 and explanation (2 marks):
Ratio NaOH : Na2SO4 = 2:1; n(Na2SO4) = 0.00900 ÷ 2 = 0.00450 mol.
V(total) = 40.0 + 60.0 = 100.0 mL = 0.1000 L; c(Na2SO4) = 0.00450 ÷ 0.1000 = 0.0450 mol/L [1].
100.0 mL is used because the Na2SO4 produced dissolves in the entire mixed solution (both original volumes combined), not just in one of the original solutions. Using 40.0 mL or 60.0 mL would give a concentration that is too high because it underestimates the volume in which the product is dissolved [1].
Q2 — Sample Band 6 response (8 marks), annotated
Error in Step 1: The student substituted 40.0 (the volume in mL) directly into n = cV without converting to litres. n = cV requires V in litres. The correct volume is 40.0 mL = 0.0400 L. Using 40.0 instead of 0.0400 gives an answer 1000× too large, which is why the student obtained 4.00 mol instead of 0.00400 mol [1].
Error in Step 2: The student used a 1:1 mole ratio, but the balanced equation shows a 2:1 ratio (2 mol NaOH reacts with 1 mol H2SO4). The correct relationship is n(H2SO4) = n(NaOH) ÷ 2 [1].
Correct working [2 marks]:
Step 1: V(NaOH) = 40.0 mL = 0.0400 L; n(NaOH) = 0.100 × 0.0400 = 0.00400 mol.
Step 2: Ratio 2NaOH : 1H2SO4; n(H2SO4) = 0.00400 ÷ 2 = 0.00200 mol.
Step 3: V(H2SO4) = 25.0 mL = 0.0250 L; c(H2SO4) = 0.00200 ÷ 0.0250 = 0.0800 mol/L [1 mark for correct calculation chain; 1 mark for correct final answer with units].
Reasonableness check [1 mark]: Concentrated H2SO4 is approximately 18 mol/L (pure acid); 160 mol/L is physically impossible because it exceeds the theoretical maximum concentration of any acid. A result more than 10× the concentration of the NaOH solution used should be immediately questioned. The correct answer of 0.0800 mol/L is a dilute solution, consistent with typical laboratory concentrations [1].
Experimental precaution [1 mark]: Accept any one of: repeat the titration at least three times and average concordant results (within 0.10 mL) to reduce random error; rinse the burette with the titrant solution before filling to avoid dilution; use a suitable indicator (e.g. phenolphthalein for strong acid/strong base) to accurately identify the endpoint; ensure no air bubbles are present in the burette tip before starting [1].
Marking criteria summary (8 marks): 1 = identifies the mL→L conversion error with explanation; 1 = identifies the incorrect mole ratio with explanation; 1 = correct n(NaOH) = 0.00400 mol shown with unit conversion; 1 = correct n(H2SO4) = 0.00200 mol using 2:1 ratio; 1 = correct c(H2SO4) = 0.0800 mol/L with units; 1 = identifies 160 mol/L as implausible using a specific physical/chemical argument; 1 = one valid experimental precaution with brief justification; 1 = uses precise chemical language throughout (mole ratio, limiting reagent/titration context, unit conversion, mol/L).