Chemistry • Year 11 • Module 2 • Lesson 15
Gas Stoichiometry
Apply the modified 4-step pathway to real gas problems, interpret data on gas volumes, and compare STP vs SATP answers in context.
1. Interpret the data table — gas volumes from three reactions
A laboratory technician recorded the gas volumes produced from three reactions at different conditions. Some cells are missing. 9 marks
| Reaction (balanced) | Reactant & mass/moles given | Conditions | Moles of gas produced | Volume of gas produced (L) |
|---|---|---|---|---|
| 2H2O2(aq) → 2H2O(l) + O2(g) | 17.0 g of H2O2 (MM = 34.015) |
STP | ||
| CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) | 12.4 L of CO2 collected at SATP | SATP (find mass of CH4 consumed) |
12.4 L (given) | |
| 2KClO3(s) → 2KCl(s) + 3O2(g) | 24.5 g of KClO3 (K=39.098, Cl=35.453, O=15.999) |
SATP |
Molar masses needed: H2O2 = 34.015; CH4 = 16.043; KClO3 = 122.55 g mol−1. Molar volumes: STP = 22.71 L mol−1; SATP = 24.8 L mol−1.
1.1 Complete the “moles of gas” and “volume of gas” columns for rows 1 and 3. For row 2, calculate the mass of CH4 consumed. Show all working below. 6 marks (2 per reaction)
1.2 For row 1, calculate what the volume of O2 would be if the conditions were SATP instead of STP. Explain in one sentence why the answer is different. 2 marks
1.3 Row 3 uses KClO3, a solid. A student tries to apply the volume ratio shortcut (V ratio = coefficient ratio) directly. Identify the error in this approach. 1 mark
2. Interpret graph — zinc metal dissolving in hydrochloric acid
A student placed 9.81 g of zinc in excess hydrochloric acid: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g). They measured the volume of H2 gas collected over the reaction (SATP conditions). The graph below shows volume collected vs time. 8 marks
Figure 2.1. Volume of H2(g) collected over time from Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g). 9.81 g Zn in excess HCl. SATP conditions. Illustrative data. (Zn = 65.38 g mol−1)
2.1 Describe the trend in volume from 0 s to 360 s. 2 marks
2.2 Calculate the theoretical volume of H2 produced at SATP from 9.81 g of Zn. Show full working and compare your answer with the plateau value on the graph. 3 marks
2.3 Explain why the graph levels off (plateau). Suggest one experimental reason the actual plateau might be slightly lower than your theoretical calculation. 2 marks
2.4 Predict the plateau volume if the same mass of Zn were reacted at STP instead of SATP. Calculate it. 1 mark
3. Compare the two approaches to gas stoichiometry
Complete the table comparing the full 4-step method (with molar volume conversion) and the volume ratio shortcut. 10 marks (1 per cell)
| Feature | Full 4-step method (+ gas conversion) | Volume ratio shortcut |
|---|---|---|
| When can it be used? | ||
| What additional data do you need? | ||
| Key formula(s) | ||
| Works when solids or liquids are involved? | ||
| Example reaction it applies to |
4. Predict and justify — industry scenario
A lime kiln in Yallourn, Victoria, heats limestone (CaCO3) to produce quicklime (CaO) and CO2 gas: CaCO3(s) → CaO(s) + CO2(g). The kiln processes 2.00 kg of pure CaCO3 per batch. (Ca = 40.078, C = 12.011, O = 15.999; molar mass of CaCO3 = 100.09 g mol−1)
5 marks
4.1 Calculate the volume of CO2 produced per batch at STP. Show full working. 3 marks
4.2 The kiln operates at high temperature, not at STP. Predict whether the actual volume of CO2 gas emitted from the kiln would be greater than, less than, or equal to your calculated value. Justify your answer in terms of the relationship between temperature, pressure and gas volume. 2 marks
Q1.1 — Data table completion
Row 1 (H2O2, STP):
n(H2O2) = 17.0 ÷ 34.015 = 0.4998 mol; ratio H2O2:O2 = 2:1; n(O2) = 0.4998 ÷ 2 = 0.2499 mol; V(O2) = 0.2499 × 22.71 = 5.67 L.
Row 2 (CH4, SATP; find m(CH4)):
n(CO2) = 12.4 ÷ 24.8 = 0.5000 mol; ratio CO2:CH4 = 1:1; n(CH4) = 0.5000 mol; m(CH4) = 0.5000 × 16.043 = 8.02 g.
Row 3 (KClO3, SATP):
MM(KClO3) = 39.098 + 35.453 + 3(15.999) = 122.55 g mol−1; n(KClO3) = 24.5 ÷ 122.55 = 0.1999 mol; ratio KClO3:O2 = 2:3; n(O2) = 0.1999 × (3 ÷ 2) = 0.2999 mol; V(O2) = 0.2999 × 24.8 = 7.44 L.
Marking: 1 mark for correct n(gas), 1 mark for correct V (per reaction). For row 2: 1 mark for correct n(CH4), 1 mark for correct m(CH4).
Q1.2 — Row 1 at SATP instead of STP
Same n(O2) = 0.2499 mol; V = 0.2499 × 24.8 = 6.20 L. The volume is larger at SATP because 25 °C is higher than 0 °C, giving gas molecules more kinetic energy and thus a larger molar volume (24.8 vs 22.71 L mol−1).
Award 1 mark for correct SATP volume; 1 mark for the explanation referencing higher temperature → larger molar volume.
Q1.3 — Error in applying volume ratio to KClO3
KClO3 and KCl are solids — they do not have gas volumes in the Avogadro sense. The volume ratio shortcut applies only when all species being compared are gases at the same T and P. Because the reactant (KClO3) is a solid, the shortcut is invalid; the full 4-step method (mass → moles → ratio → volume) must be used.
Q2.1 — Trend (2 marks)
The volume of H2 rises steeply from 0 mL at time zero, with the rate of increase slowing from about 180 s onward [1]. By approximately 240–300 s the volume levels off at a plateau of about 3720 mL as the zinc is fully consumed [1].
Q2.2 — Theoretical volume (3 marks)
n(Zn) = 9.81 ÷ 65.38 = 0.1500 mol [1]; ratio Zn:H2 = 1:1; n(H2) = 0.1500 mol [1]; V(H2) = 0.1500 × 24.8 = 3.72 L = 3720 mL at SATP [1]. This matches the plateau on the graph closely, confirming the calculation.
Q2.3 — Plateau explanation (2 marks)
The graph levels off because the zinc (limiting reagent) has been completely consumed; no more H2 can be produced [1]. The actual plateau might be slightly lower than the theoretical value because some gas could dissolve in the acidic solution, or because a small amount of Zn may have impurities (not pure Zn), reducing the effective mass reacting [1]. Accept: gas loss before the collection tube sealed; small experimental error in mass measurement.
Q2.4 — Volume at STP (1 mark)
Same n(H2) = 0.1500 mol; V(H2) at STP = 0.1500 × 22.71 = 3.41 L = 3410 mL [1]. This is lower than the SATP value because 0 °C < 25 °C and the gas occupies less volume at the lower temperature.
Q3 — Compare and contrast table
When can it be used? Full: always, for any stoichiometry problem involving a gas. Shortcut: only when all species being compared are gases at the same T and P.
Additional data needed? Full: the molar volume for the stated conditions (22.71 or 24.8 L mol−1), plus molar mass if a solid/liquid mass is given. Shortcut: no molar volume needed — just the balanced equation coefficients.
Key formulas? Full: n = m ÷ MM; mole ratio; V = n × Vm (or n = V ÷ Vm). Shortcut: VA / VB = coefficientA / coefficientB.
Works with solids/liquids? Full: Yes. Shortcut: No.
Example reaction? Full: CaCO3(s) → CaO(s) + CO2(g); any reaction with a solid or liquid. Shortcut: H2(g) + Cl2(g) → 2HCl(g); N2(g) + 3H2(g) → 2NH3(g).
Award 1 mark per correct cell (both cells must be correct for that row to receive 2 marks). Accept reasonable variations.
Q4.1 — Lime kiln CO2 at STP (3 marks)
n(CaCO3) = 2000 ÷ 100.09 = 19.98 mol [1]; ratio CaCO3:CO2 = 1:1; n(CO2) = 19.98 mol [1]; V(CO2) = 19.98 × 22.71 = 453.6 L ≈ 454 L at STP [1].
Q4.2 — Kiln temperature prediction (2 marks)
The actual volume of CO2 emitted from the kiln would be much greater than the calculated STP value [1]. At higher temperatures (lime kilns operate above 900 °C), the gas molecules have far greater kinetic energy and occupy a much larger volume at the same pressure; Charles’s law states that volume is directly proportional to absolute temperature at constant pressure, so a 900 °C kiln would produce CO2 at a volume roughly 4× larger than at 0 °C (1173 K ÷ 273 K) [1].