Chemistry • Year 11 • Module 2 • Lesson 12
Mass–Mass Stoichiometry
Build HSC Band 5–6 technique: multi-part industrial calculations, conservation of mass verification, and evaluating why the 4-step method is the only reliable approach.
1. Multi-part investigation — Haber process
10 marks Band 4–5
Context. The Haber process is an industrial reaction used in Australia’s fertiliser industry. It combines nitrogen and hydrogen under high pressure and temperature to produce ammonia:
N2(g) + 3H2(g) ⇌ 2NH3(g)
A production batch starts with 28.0 g of N2 and an unlimited supply of H2. Assume complete conversion of N2 to NH3. (N = 14.007, H = 1.008)
(a) Show all four steps to calculate the mass of NH3 produced. 4 marks
(b) Calculate the mass of H2 consumed in the same reaction. 3 marks
(c) Verify conservation of mass by comparing the total mass of reactants used with the total mass of NH3 produced. Is mass conserved? Explain why or why not. 3 marks
2. Thermite reaction — evaluate the 4-step method
8 marks Band 4–5
Context. The thermite reaction is used in railway track welding in Australia:
2Al + Fe2O3 → Al2O3 + 2Fe
A rail contractor uses 108 g of aluminium powder for each weld. (Al = 26.982, Fe = 55.845, O = 15.999)
(a) Calculate the mass of Fe2O3 that reacts completely with 108 g of Al. Show all four steps. 4 marks
(b) Calculate the theoretical mass of iron (Fe) produced from 108 g of Al. 2 marks
(c) A student claims: “I can find the mass of Al2O3 produced simply by subtracting the mass of Fe from the mass of Fe2O3 reacted.” Is this reasoning correct? Explain with reference to conservation of mass. 2 marks
3. Extended analysis — evaluate a flawed shortcut (Band 5–6)
7 marks Band 5–6
Scenario. Two students calculate the mass of SO3 produced when 12.8 g of SO2 reacts in the Contact process:
2SO2(g) + O2(g) → 2SO3(g). (S = 32.06, O = 15.999)
Student A uses the 4-step method and obtains 16.0 g.
Student B says: “The ratio is 2:2 = 1:1, so n(SO3) = n(SO2), and I can skip the ratio step and just use the molar masses directly.” Student B also obtains 16.0 g.
Q3. Evaluate Student B’s approach. In your response you must:
- Verify Student A’s answer by showing the full 4-step working.
- Explain whether Student B’s answer is correct by coincidence or by valid method, and why.
- Identify a different reaction where Student B’s shortcut would give the wrong answer, and calculate the correct answer using the 4-step method to show the error.
- State the general principle that determines when a 1:1 mole ratio can be assumed.
Q1(a) — Mass of NH3 produced (4 marks)
Step 1: Balanced. Ratio N2 : NH3 = 1 : 2. [1 mark]
Step 2: MM(N2) = 2(14.007) = 28.014; n(N2) = 28.0 ÷ 28.014 = 0.9995 mol. [1 mark]
Step 3: n(NH3) = 0.9995 × 2 ÷ 1 = 1.999 mol. [1 mark]
Step 4: MM(NH3) = 14.007 + 3(1.008) = 17.031; m(NH3) = 1.999 × 17.031 = 34.0 g. [1 mark]
Marking criteria: 1 each for correct ratio; n(N2); n(NH3); m(NH3) with MM shown.
Q1(b) — Mass of H2 consumed (3 marks)
Ratio N2 : H2 = 1 : 3. [1 mark] n(H2) = 0.9995 × 3 = 2.999 mol. [1 mark] MM(H2) = 2(1.008) = 2.016; m(H2) = 2.999 × 2.016 = 6.05 g. [1 mark]
Q1(c) — Conservation of mass verification (3 marks)
Total reactant mass = m(N2) + m(H2) = 28.0 + 6.05 = 34.05 g ≈ 34.0 g. [1 mark]
Total product mass = m(NH3) = 34.0 g. [1 mark]
Reactant total ≈ product total ⇒ mass is conserved. [1 mark] This is consistent with the Law of Conservation of Mass: atoms are rearranged from N2 and H2 into NH3; no atoms are created or destroyed, so the total mass before and after reaction is equal.
Small rounding differences (<0.1 g) are acceptable. Award mark if student explains mass conservation in terms of atom rearrangement.
Q2(a) — Mass of Fe2O3 (4 marks)
Step 1: Ratio Al : Fe2O3 = 2 : 1. [1 mark]
Step 2: n(Al) = 108 ÷ 26.982 = 4.003 mol. [1 mark]
Step 3: n(Fe2O3) = 4.003 × 1 ÷ 2 = 2.002 mol. [1 mark]
Step 4: MM(Fe2O3) = 2(55.845) + 3(15.999) = 159.69; m(Fe2O3) = 2.002 × 159.69 = 319.7 g ≈ 320 g. [1 mark]
Q2(b) — Mass of Fe produced (2 marks)
Ratio Al : Fe = 2 : 2 = 1 : 1. n(Fe) = 4.003 mol (same as n(Al) via 1:1 ratio). [1 mark] m(Fe) = 4.003 × 55.845 = 223.5 g ≈ 224 g. [1 mark]
Q2(c) — Evaluate student’s shortcut (2 marks)
The reasoning is incorrect. [1 mark] The student assumes that all the mass of Fe2O3 is converted into Fe, but this is not the case. The oxygen atoms in Fe2O3 end up bonded to aluminium in Al2O3, not in Fe. Conservation of mass applies to the entire system — total reactant mass (108 g Al + 320 g Fe2O3 = 428 g) equals total product mass (m(Al2O3) + m(Fe)). The mass of Al2O3 must be calculated using the 4-step method with the Al:Al2O3 ratio, not by subtraction of Fe from Fe2O3. [1 mark]
Q3 — Band 5–6 extended answer (7 marks), annotated
Verification of Student A (4 steps): Ratio SO2 : SO3 = 2 : 2 = 1 : 1 [1]. MM(SO2) = 32.06 + 2(15.999) = 64.058; n(SO2) = 12.8 ÷ 64.058 = 0.1998 mol [1]. n(SO3) = 0.1998 × 1 = 0.1998 mol [ratio step has no numerical effect since 1:1] [1]. MM(SO3) = 32.06 + 3(15.999) = 80.057; m(SO3) = 0.1998 × 80.057 = 16.0 g [1].
Student B’s answer is correct by coincidence [1]: Because the mole ratio SO2 : SO3 = 2 : 2 = 1 : 1, applying the ratio multiplies n(SO2) by 1 — it has no numerical effect. Student B’s shortcut works here only because the ratio is 1:1. This is not a valid general method; it is a conceptual error that happens to give the right number in this specific case.
Counter-example showing the shortcut fails [1]: Consider 4Al + 3O2 → 2Al2O3. If 27.0 g of Al reacts, the 4-step method gives: ratio Al : Al2O3 = 4 : 2 = 2 : 1; n(Al) = 27.0 ÷ 26.982 = 1.001 mol; n(Al2O3) = 1.001 × 2 ÷ 4 = 0.5003 mol; m(Al2O3) = 0.5003 × 101.96 = 51.0 g. Student B’s shortcut (using n(Al2O3) = n(Al)) gives m(Al2O3) = 1.001 × 101.96 = 102.1 g — double the correct answer.
General principle [1]: A 1:1 mole ratio can only be assumed when the two species have equal coefficients in the balanced equation (e.g. both are 2, both are 3, etc.). The ratio must always be read from the balanced equation; it cannot be assumed.
Marking criteria: 1 mark × 4 for full 4-step verification; 1 mark for correctly labelling B’s result as coincidental with explanation; 1 mark for a valid counter-example with 4-step calculation showing the error; 1 mark for a correctly stated general principle.