HSC Exam Practice

Sample response. Solution stoichiometry refers to calculations that determine the quantity of a product (mass or concentration) from the concentration and volume of one or more reactant solutions, using the mole ratios from a balanced equation. It combines (1) the concentration/volume/moles relationship (IQ3: n = cV) with (2) the four-step stoichiometric method using balanced equations and mole ratios (IQ1).

Marking notes. 1 mark for a correct definition of solution stoichiometry (must mention both “concentration and volume” and “mole ratios/stoichiometry”); 1 mark for identifying the first sub-skill (n = cV / concentration and volume gives moles); 1 mark for identifying the second sub-skill (balanced equation / mole ratio from stoichiometry).

Sample response. The equation n = cV is derived from the definition of concentration: c = n/V, where V must be expressed in litres so that the units are consistent (mol L⁻¹ × L = mol). If 25.0 mL is used directly (without converting), the calculation gives n = c × 25.0, which is 1000× too large because 25.0 mL = 0.0250 L and 25.0 ÷ 0.0250 = 1000. The mass or concentration of product calculated would therefore be 1000× larger than the correct value.

Marking notes. 1 mark for explaining that c is defined in mol L⁻¹ so V must be in litres for unit consistency; 1 mark for stating that not converting gives an answer 1000× too large; 1 mark for explaining that 25.0 mL = 0.0250 L (or that dividing by 1000 is required) and the consequence for the final answer.

Sample response. Step 0 (prerequisite): write and balance the chemical equation; extract the mole ratio and convert mL to L. Step 1: calculate moles of the given reactant using n = c × V (V must be in litres). Step 2: apply the mole ratio from the balanced equation to find moles of the desired product: n(product) = n(reactant) × [coefficient of product ÷ coefficient of reactant]. Step 3: convert moles of product to mass using m = n × MM, where MM is calculated from the periodic table.

Marking notes. 1 mark per correctly stated and labelled step with the appropriate formula. Step 0/unit conversion may be combined with Step 1 — award the mark if the unit conversion is explicitly stated at any point.

Sample response. To calculate n(reactant), use only the individual volume of that reactant’s solution (V₁ for reactant A, V₂ for reactant B). Each reactant is calculated independently: n_A = c_A × V₁. To calculate c(product) in the final mixed solution, use the total combined volume V(total) = V₁ + V₂, because the product dissolves throughout the entire volume of the mixture. Using the wrong volume in either step would give an incorrect answer.

Marking notes. 1 mark for stating that n(reactant) requires only the individual solution’s volume; 1 mark for stating that c(product) requires V₁ + V₂ (total combined volume); 1 mark for a clear explanation of why (product distributes through the total volume / individual volumes belong to individual reactants).

Sample response. Calculate n = cV for each reactant separately using their individual volumes (converted to litres). Divide each n value by the stoichiometric coefficient of that species in the balanced equation. The reactant with the smaller (n ÷ coefficient) value is the limiting reagent — it will be completely consumed first. All subsequent yield calculations must be based on the amount of limiting reagent present, not the excess reactant.

Marking notes. 1 mark for stating that n = cV is calculated independently for each reactant; 1 mark for stating the comparison step (n ÷ coefficient for each); 1 mark for correctly identifying that the smaller value indicates the limiting reagent.

Sample response. Using V₁ + V₂ to calculate n(reactant A) is incorrect because reactant A was only dissolved in volume V₁, not in the combined volume. Substituting V(total) inflates the calculated n value: n = c_A × (V₁ + V₂) > c_A × V₁. This incorrectly implies that more moles of reactant are present than actually exist in the solution, leading to an overestimate of the moles of precipitate produced, and therefore an overestimate of its mass. The fundamental error is treating the reactant as if it were dissolved in the entire combined volume rather than its own original volume.

Marking notes. 1 mark for identifying that using V₁ + V₂ for a reactant that was only in V₁ is conceptually incorrect; 1 mark for stating that this inflates n(reactant); 1 mark for explaining the downstream consequence (overestimate of mass of precipitate or moles of product).

Part (a) — n for each reactant (2 marks). V(Pb(NO₃)₂) = 30.0 mL = 0.0300 L; n(Pb(NO₃)₂) = 0.200 × 0.0300 = 0.00600 mol [1]. V(KI) = 40.0 mL = 0.0400 L; n(KI) = 0.250 × 0.0400 = 0.01000 mol [1].

Part (b) — Limiting reagent (2 marks). Pb(NO₃)₂: 0.00600 ÷ 1 = 0.00600; KI: 0.01000 ÷ 2 = 0.00500 [1]. KI gives the smaller value (0.00500 < 0.00600), so KI is the limiting reagent and Pb(NO₃)₂ is in excess [1].

Part (c) — Mass of PbI₂ (2 marks). Ratio KI : PbI₂ = 2:1; n(PbI₂) = 0.01000 ÷ 2 = 0.00500 mol [1]. MM(PbI₂) = 207.2 + 2(126.90) = 461.0 g/mol; m(PbI₂) = 0.00500 × 461.0 = 2.31 g [1].

Part (d) — Moles of excess reagent remaining (2 marks). Excess reagent is Pb(NO₃)₂. Ratio KI : Pb(NO₃)₂ = 2:1; n(Pb(NO₃)₂ consumed) = n(KI) × (1/2) = 0.01000 × 0.500 = 0.00500 mol [1]. n(Pb(NO₃)₂ remaining) = 0.00600 − 0.00500 = 0.00100 mol [1].

Sample response. The claim has merit but requires qualification. The mL→L error produces a catastrophic numerical error (factor of 1000), while a mole-ratio error produces a smaller but still serious proportional error. Both types cause incorrect answers, but their impact and detectability differ. Consider the calculation: 25.0 mL of 0.100 mol/L AgNO₃ reacting 1:1 with excess NaCl to form AgCl (MM = 143.32 g/mol). Correct answer: n = 0.100 × 0.0250 = 0.00250 mol; m = 0.00250 × 143.32 = 0.358 g. Error 1 (mL not converted): n = 0.100 × 25.0 = 2.50 mol; m = 2.50 × 143.32 = 358 g — a result 1000× too large. For a typical laboratory reaction this is physically implausible (several kilograms of precipitate from a small beaker), making the error detectable if a reasonableness check is applied. Error 2 (mole ratio applied as 1:2 instead of 1:1): n(AgCl) = 0.00250 ÷ 2 = 0.00125 mol; m = 0.00125 × 143.32 = 0.179 g — a result 2× too small. This error is harder to detect by inspection because 0.179 g is plausible. The claim that the unit error is “more common” is supported by the lesson’s Common Mistakes section, which lists it first. However, the mole-ratio error is arguably more insidious because the incorrect answer still looks numerically reasonable. Strategies: (1) always write “V = ___ mL = ___ L” as the first line of every solution; (2) circle the mole ratio in the equation before calculating; (3) perform a reasonableness check — compare your answer to typical laboratory scales (grams, not kilograms). Together these habits eliminate both error types under exam conditions.

Marking criteria (6 marks). 1 = correctly calculates and states the impact of the mL→L error on a specific numerical example (factor of 1000 or ×1000 in answer). 1 = correctly calculates and states the impact of a mole-ratio error on a specific numerical example (factor proportional to ratio error). 1 = compares the two errors (e.g. mL error is catastrophic/easily detectable; ratio error is proportional/subtle). 1 = reaches an evaluative judgement about the claim (agrees with qualification, or disagrees with justification, explicitly weighing severity and frequency). 1 = identifies at least one specific, actionable strategy to prevent the mL→L error in an exam. 1 = identifies at least one specific, actionable strategy to prevent a mole-ratio error in an exam.