Chemistry • Year 11 • Module 2 • Lesson 11
Stoichiometry — Mole Ratios
Apply your understanding of mole ratios, balancing equations and stoichiometric calculations to real data, industrial contexts and scenario reasoning.
1. Interpret stoichiometric data — Haber process and related reactions
The table below lists four balanced equations used in industrial and laboratory chemistry in Australia. 9 marks
| Equation | Balanced chemical equation | Given moles | Find n(species) | Calculated moles |
|---|---|---|---|---|
| A | 2Mg + O2 → 2MgO | 3.00 mol Mg | n(MgO) | |
| B | 2Mg + O2 → 2MgO | 1.50 mol O2 | n(MgO) | |
| C | N2 + 3H2 → 2NH3 | 2.00 mol N2 | n(H2) | |
| D | N2 + 3H2 → 2NH3 | 5.00 mol NH3 | n(N2) | |
| E | 3Fe3O4 + 8Al → 9Fe + 4Al2O3 | 1.20 mol Fe3O4 | n(Al2O3) | |
| F | 3Fe3O4 + 8Al → 9Fe + 4Al2O3 | 0.600 mol Al | n(Fe) |
1.1 Complete the “Calculated moles” column. Show the mole-ratio formula you used for each row. 6 marks (1 per row)
1.2 For equation C, a student incorrectly used the subscript 2 from N2 instead of the coefficient 1. What incorrect answer would they get for n(H2)? Explain why this is wrong. 2 marks
1.3 Rows E and F use the thermite reaction (used in railway weld repair in Australia). If a rail welding kit provides 2.40 mol Fe3O4 and 6.40 mol Al, identify which reactant limits the yield of Fe and explain your reasoning using mole ratios. 1 mark
2. Interpret graph — Haber process NH3 production
A Year 11 class modelled a scaled Haber process, starting with varying amounts of N2 and always using the correct stoichiometric amount of H2. The graph below shows theoretical n(NH3) produced as a function of n(N2) used. 7 marks
Figure 2.1. Theoretical n(NH3) produced vs n(N2) used in N2 + 3H2 → 2NH3, with stoichiometric H2. Illustrative data.
2.1 Describe the relationship shown in the graph between n(N2) and n(NH3). Link this relationship to the mole ratio from the balanced equation. 2 marks
2.2 Use the graph to estimate n(NH3) produced when 3.5 mol N2 is used. Confirm your estimate using the mole-ratio formula. 2 marks
2.3 A student claims that because the graph passes through the origin, there is a direct proportion between n(N2) and n(NH3), and therefore the gradient equals the mole ratio N2 : NH3. Evaluate this claim — is it correct? What is the gradient and what does it represent? 3 marks
3. Compare coefficient vs subscript across five features
Complete the two-column table. For each feature, write a concise description that distinguishes a coefficient from a subscript. 10 marks (1 per cell)
| Feature | Coefficient (large number, in front of formula) | Subscript (small number, inside formula) |
|---|---|---|
| What it counts | ||
| Can be changed when balancing? | ||
| Position in equation | ||
| Used to determine mole ratio? | ||
| Example from 2H2O |
4. Predict and justify — methane combustion in a Pilbara gas plant
A natural-gas processing plant in Western Australia burns methane (CH4) as a fuel according to the equation: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g). An engineer has 120 mol of CH4 available and an unlimited supply of O2.
5 marks
4.1 Predict how many moles of CO2 and H2O would be produced if all 120 mol CH4 burns completely. Show the ratio you used. 2 marks
4.2 The plant engineer claims: “Because CH4 and CO2 have a 1 : 1 mole ratio, and CH4 and H2O also have a 1 : 2 ratio, all combustion reactions must have at least one product in excess of 1.” Evaluate this claim. Is it always true? Give a counterexample or justify why it holds for all combustion reactions of hydrocarbons. 3 marks
5. Case study — iron production at Port Kembla
The Port Kembla steelworks in NSW uses a blast furnace to reduce iron oxide (Fe2O3) with carbon monoxide according to the equation: Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g). In 2022, the furnace processed approximately 1200 mol of Fe2O3 per batch. 4 marks
5.1 Calculate the moles of CO required and the moles of Fe and CO2 theoretically produced per batch. Show all mole-ratio working. 3 marks
5.2 The actual yield of Fe per batch is 1920 mol, not the theoretical value. Calculate the percentage yield and suggest one reason why the actual yield is less than theoretical. 1 mark
Q1.1 — Calculated moles
A: n(MgO) = 3.00 × (2÷2) = 3.00 mol. B: n(MgO) = 1.50 × (2÷1) = 3.00 mol. C: n(H2) = 2.00 × (3÷1) = 6.00 mol. D: n(N2) = 5.00 × (1÷2) = 2.50 mol. E: n(Al2O3) = 1.20 × (4÷3) = 1.60 mol. F: n(Fe) = 0.600 × (9÷8) = 0.675 mol.
Q1.2 — Subscript error
Using the subscript 2 from N2 instead of the coefficient 1 would give n(H2) = 2.00 × (3÷2) = 3.00 mol, which is wrong. The correct answer is 6.00 mol. The subscript 2 in N2 tells you there are two nitrogen atoms in one nitrogen molecule — it has nothing to do with how many moles of N2 react.
Q1.3 — Limiting reactant (thermite)
From the equation 3Fe3O4 + 8Al → …, the mole ratio Fe3O4 : Al = 3 : 8. For 2.40 mol Fe3O4, the stoichiometric amount of Al required = 2.40 × (8÷3) = 6.40 mol. We have exactly 6.40 mol Al — neither reactant is in excess; both are consumed completely. If any less Al were available, Al would be the limiting reactant.
Q2.1 — Graph relationship and mole ratio (2 marks)
The graph shows a directly proportional (linear through origin) relationship between n(N2) and n(NH3) [1]. This reflects the 1 : 2 mole ratio in the balanced equation N2 + 3H2 → 2NH3: for every 1 mol N2 consumed, 2 mol NH3 is produced [1].
Q2.2 — Estimate and confirm (2 marks)
From graph at x = 3.5: y ≈ 7 mol NH3 [1]. Confirmed by formula: n(NH3) = 3.5 × (2÷1) = 7.0 mol NH3 [1].
Q2.3 — Gradient claim (3 marks)
The claim is largely correct [1]. The graph does pass through the origin confirming direct proportion. The gradient = Δn(NH3) / Δn(N2) = 2 mol NH3 / 1 mol N2 = 2 [1]. This gradient equals the ratio coeff(NH3) ÷ coeff(N2) = 2÷1 = 2, which is a component of the mole ratio. Strictly, “the mole ratio” N2 : NH3 = 1 : 2, and the gradient represents the factor by which n(NH3) exceeds n(N2) — not the full ratio. The statement is functionally correct but needs the clarification that the gradient = coeff(wanted) / coeff(given), not the ratio itself [1].
Q3 — Compare coefficient vs subscript
What it counts: Coefficient: number of moles of an entire formula unit in a reaction. Subscript: number of atoms of one element within a single formula unit. Changed when balancing? Coefficient: Yes — changing a coefficient is the only permitted way to balance. Subscript: No — changing a subscript changes the identity of the substance. Position: Coefficient: in front of (before) the chemical formula, large number. Subscript: inside the formula, small number written below the element symbol. Used for mole ratio? Coefficient: Yes — the mole ratio is read directly from the coefficients. Subscript: No. Example 2H2O: Coefficient: 2 (means 2 mol H2O are involved). Subscript: 2 in H2 (means 2 hydrogen atoms per water molecule).
Q4.1 — CH4 combustion products (2 marks)
Ratio CH4 : CO2 = 1 : 1. n(CO2) = 120 × (1÷1) = 120 mol. Ratio CH4 : H2O = 1 : 2. n(H2O) = 120 × (2÷1) = 240 mol [1 mark for each product].
Q4.2 — Combustion ratio claim (3 marks)
The engineer’s claim is not universally true [1]. For CH4 + 2O2 → CO2 + 2H2O, CO2 is 1:1 and H2O is 1:2 — H2O is in excess of 1 per mole CH4. For a hydrocarbon CxHy, the H2O coefficient is y/2 and CO2 coefficient is x; if x = y/2 (e.g. CH2, a fragment not a real fuel), both would be equal. For carbon (C) combustion (C + O2 → CO2), the ratio is 1:1:1 — no product exceeds the fuel in mole ratio [1]. Therefore the claim does not hold for all combustion reactions; it only holds for hydrocarbons where the hydrogen count causes H2O production to exceed 1 mol per mol fuel [1].
Q5.1 — Blast furnace mole ratios (3 marks)
Ratio Fe2O3 : CO = 1 : 3. n(CO) = 1200 × 3 = 3600 mol [1].
Ratio Fe2O3 : Fe = 1 : 2. n(Fe) = 1200 × 2 = 2400 mol [1].
Ratio Fe2O3 : CO2 = 1 : 3. n(CO2) = 1200 × 3 = 3600 mol [1].
Q5.2 — Percentage yield
% yield = (1920 ÷ 2400) × 100 = 80%. One possible reason: not all Fe2O3 reacted completely (incomplete conversion); or side reactions produced other iron compounds (e.g. Fe3O4); or product loss during tapping and processing [any one valid reason].