Chemistry • Year 11 • Module 2 • Lesson 11
Stoichiometry — Mole Ratios
Build HSC Band 5–6 extended-response technique on mole-ratio reasoning, multi-step stoichiometric calculations, and evaluating claims using balanced equations in industrial contexts.
1. Data + scenario: the Haber process at Gibson Island (Band 5–6)
8 marks Band 5–6
Scenario. The former Incitec Pivot ammonia plant at Gibson Island, Brisbane, used the Haber process: N2(g) + 3H2(g) → 2NH3(g). In a single production run, a process engineer recorded the following data before and after a reactor cycle.
| Measurement | Before reaction | After reaction |
|---|---|---|
| n(N2) in reactor (mol) | 800 | 200 |
| n(H2) in reactor (mol) | 2400 | 600 |
| n(NH3) in reactor (mol) | 0 | 1200 |
| Total moles of gas in reactor | 3200 | 2000 |
Illustrative data based on typical Haber process single-pass conversion rates (~37.5% under high pressure). Gibson Island plant operated 1968–2022.
Q1. Analyse and evaluate the experimental data above to assess whether the reactor data are consistent with the balanced equation for the Haber process. In your response you must:
- Verify that the change in moles of N2, H2 and NH3 is consistent with the mole ratio in the balanced equation, showing quantitative working.
- Explain why the total moles of gas in the reactor decreased even though the Law of Conservation of Mass holds.
- Calculate the single-pass conversion percentage for N2 and explain what this means for process efficiency.
- Predict the effect on n(NH3) produced if the engineer had started with 800 mol N2 but only 2000 mol H2 (instead of 2400 mol). Identify which reactant becomes limiting and calculate the maximum n(NH3) possible.
- State one industrial reason why the Haber process is run at less than 100% conversion per pass, with reference to economic efficiency.
2. Multi-step stoichiometric calculation — propane combustion (Band 5–6)
7 marks Band 5–6
Context. Propane (C3H8) is used as a fuel for remote area generators in outback Australia. The complete combustion equation is: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g).
Q2(a). A generator burns 4.50 mol of C3H8 per hour in a complete combustion. Calculate (i) n(O2) consumed per hour and (ii) n(CO2) and n(H2O) produced per hour. Show mole ratio working for each species. 3 marks (1 per species)
Q2(b). The supply cylinder contains 18.0 mol of C3H8. The generator is also connected to an oxygen cylinder containing 80.0 mol O2. Determine which reactant is the limiting reagent and calculate the maximum moles of CO2 that can be produced before the limiting reagent is exhausted. 2 marks
Q2(c). Verify that the Law of Conservation of Mass holds for the complete combustion of 18.0 mol C3H8 using the theoretical moles from part (a) (with O2 in excess). Calculate the total mass of reactants (C3H8 + O2) and compare it to the total mass of products (CO2 + H2O). (C=12.011, H=1.008, O=15.999) 2 marks
Q1 — Sample Band 6 response (8 marks), annotated
Mole ratio verification: From the data, Δn(N2) consumed = 800 − 200 = 600 mol; Δn(H2) consumed = 2400 − 600 = 1800 mol; Δn(NH3) produced = 1200 − 0 = 1200 mol [working shown]. The ratio of changes is 600 : 1800 : 1200 = 1 : 3 : 2, which exactly matches the coefficients in N2 + 3H2 → 2NH3. The data are fully consistent with the balanced equation [1 mark].
Why total moles decreased: The Law of Conservation of Mass means atoms are rearranged, not created or destroyed — total mass is conserved. However, the number of moles of gas is not necessarily conserved. In this reaction, 4 mol of gas (1 N2 + 3 H2) combine to produce only 2 mol of gas (2 NH3). The net decrease is 2 mol per reaction unit. Total moles decreased from 3200 to 2000 (a decrease of 1200 mol), which corresponds to 600 reaction units each consuming a net 2 mol of gas [1 mark]. Mass is still conserved because the molecular mass of 2 mol NH3 (2 × 17 = 34 g) equals the mass of 1 mol N2 (28 g) + 3 mol H2 (6 g) = 34 g [1 mark].
Single-pass conversion of N2: N2 consumed = 600 mol; initial N2 = 800 mol. % conversion = (600 / 800) × 100 = 75% [1 mark]. This means 25% of the N2 (200 mol) is unreacted after each pass, limiting the yield of ammonia to 75% of the theoretical maximum per pass.
Limiting reactant prediction (800 mol N2, 2000 mol H2): Stoichiometric H2 required for 800 mol N2 = 800 × 3 = 2400 mol. We only have 2000 mol H2 < 2400 mol required. H2 is the limiting reactant [1 mark]. Maximum n(NH3) = 2000 × (2/3) = 1333 mol (not 1200 mol as in the original run, because in this scenario H2 runs out before all N2 is consumed) [1 mark].
Industrial reason for less than 100% conversion: Operating at 100% conversion per pass is impractical because it would require very high pressures and long residence times, both of which are expensive and raise safety concerns. Unreacted N2 and H2 are recycled back through the reactor, so overall conversion approaches close to 100% economically by reusing feedstock without the capital cost of a single-pass complete-conversion reactor [1 mark].
Marking criteria summary (8 marks): [1] Quantitative ratio check (600:1800:1200 = 1:3:2 matches equation). [1] Explains moles decrease (4 mol gas → 2 mol gas per unit). [1] Conservation of mass still holds (mass in = mass out, quantitative). [1] % conversion = 75% with working. [1] Identifies H2 as limiting when only 2000 mol available. [1] Calculates n(NH3) = 1333 mol (or 1333.3 mol) for the limiting scenario. [1] States a valid industrial reason for sub-100% conversion. [1] Uses precise chemical terminology throughout (mole ratio, limiting reactant, conservation of mass, coefficient, single-pass conversion).
Q2(a) — Propane combustion products (3 marks)
Ratio C3H8 : O2 = 1 : 5. n(O2) = 4.50 × 5 = 22.5 mol per hour [1].
Ratio C3H8 : CO2 = 1 : 3. n(CO2) = 4.50 × 3 = 13.5 mol per hour [1].
Ratio C3H8 : H2O = 1 : 4. n(H2O) = 4.50 × 4 = 18.0 mol per hour [1].
Q2(b) — Limiting reactant and max CO2 (2 marks)
O2 required for 18.0 mol C3H8 = 18.0 × 5 = 90.0 mol. Available O2 = 80.0 mol < 90.0 mol required. Therefore O2 is the limiting reactant [1].
n(CO2) based on limiting O2: ratio O2 : CO2 = 5 : 3. n(CO2) = 80.0 × (3/5) = 48.0 mol [1].
Q2(c) — Conservation of mass verification (2 marks)
For 18.0 mol C3H8 reacting with 90.0 mol O2 (O2 in excess, using theoretical ratios from part a):
Reactants: m(C3H8) = 18.0 × 44.097 = 793.7 g; m(O2) = 90.0 × 31.998 = 2879.8 g. Total reactant mass = 793.7 + 2879.8 = 3673.5 g [1].
Products: n(CO2) = 54.0 mol, m(CO2) = 54.0 × 44.010 = 2376.5 g; n(H2O) = 72.0 mol, m(H2O) = 72.0 × 18.015 = 1297.1 g. Total product mass = 2376.5 + 1297.1 = 3673.6 g [1]. The total mass of reactants equals the total mass of products (within rounding), confirming the Law of Conservation of Mass: atoms are rearranged but not created or destroyed.