Gas stoichiometry is not a new method — it is the 4-step method with one extra conversion step added. When a gas is given or asked for, you convert between volume and moles using molar volume, then proceed as normal. The only trap is choosing the right molar volume for the stated conditions.
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Imagine you burn a piece of magnesium ribbon in oxygen: 2Mg + O₂ → 2MgO. If you know the mass of magnesium used, what extra step would you need to find the volume of oxygen gas consumed — and why can't you just use the same 4-step method you already know?
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📚 Core Content
Wrong: Concentration and amount of solute are the same thing.
Right: Concentration is amount per unit volume; the same amount of solute can produce different concentrations in different volumes.
Avogadro's law states that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. This means 1 mole of any gas — regardless of what it is — occupies the same volume under the same conditions.
The 4-step stoichiometry method gains one extra step when a gas is involved. If gas volume is given, add a step before Step 2 to convert V → n. If gas volume is the answer, add a step after Step 3 to convert n → V.
🧮 Worked Examples
🧪 Activities
1 What volume of CO₂ at RTP is produced when 10.0 g of C burns completely? C + O₂ → CO₂. (C = 12.011)
2 What mass of Na₂O forms when 5.67 L of O₂ at STP reacts with excess Na? 4Na + O₂ → 2Na₂O. (Na = 22.990, O = 15.999)
3 What volume of H₂ at RTP is needed to reduce 8.00 g of CuO? CuO + H₂ → Cu + H₂O. (Cu = 63.546, O = 15.999)
4 In N₂ + 3H₂ → 2NH₃ at constant T and P, 6.00 L of H₂ reacts completely. What volume of NH₃ is produced?
5 What mass of Fe₂O₃ is needed to produce 11.2 L of O₂ at STP from the decomposition: 6Fe₂O₃ → 4Fe₃O₄ + O₂? (Fe = 55.845, O = 15.999)
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| Reaction | Given | Find | Conditions |
|---|---|---|---|
| 2H₂O₂ → 2H₂O + O₂ | m(H₂O₂) = 17.0 g | V(O₂) = ? | STP |
| CH₄ + 2O₂ → CO₂ + 2H₂O | V(CO₂) = 12.4 L | m(CH₄) used = ? | RTP |
| 2KClO₃ → 2KCl + 3O₂ | m(KClO₃) = 24.5 g | V(O₂) = ? | RTP |
H=1.008, O=15.999, C=12.011, K=39.098, Cl=35.453
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At the start of this lesson, you thought about what extra step is needed to find the volume of gas consumed or produced in a stoichiometry problem.
The answer is: gas stoichiometry is simply the 4-step method with one extra conversion. Before Step 1 (if gas volume is given), use n = V ÷ molar volume to convert to moles. After Step 3 (if gas volume is the answer), use V = n × molar volume. The molar volume is 22.71 L/mol at STP (0°C) or 24.8 L/mol at RTP (25°C) — always read the conditions in the question before choosing.
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✍️ Short Answer
6. When limestone (CaCO₃) is heated in a kiln, it decomposes: CaCO₃ → CaO + CO₂. A kiln processes 500 kg of limestone per hour. (a) Calculate the volume of CO₂ produced per hour at STP. (b) Explain why the actual volume produced would differ from your calculated value. (Ca=40.078, C=12.011, O=15.999) 5 MARKS
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7. 6.54 g of zinc reacts with excess hydrochloric acid: Zn + 2HCl → ZnCl₂ + H₂. (a) Calculate the volume of H₂ produced at RTP. (b) A student accidentally uses the STP molar volume. Calculate their answer and the percentage error this introduces. (Zn=65.38) 5 MARKS
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8. Explain why the volume ratio shortcut (volume ratio = coefficient ratio) can be applied to the reaction H₂ + Cl₂ → 2HCl but cannot be applied to CaCO₃ → CaO + CO₂ when finding the volume of CO₂ from a given mass of CaCO₃. 3 MARKS
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9. Consider the combustion of propane: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g). A student calculates that burning 44.1 g of propane at RTP produces 66.7 L of CO₂. A second student using the volume ratio shortcut from the equation says the answer is 66.5 L. (a) Identify which method each student used. (b) Explain any difference in the results, and state which approach is valid and why. (C=12.011, H=1.008) 4 MARKS
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10. A chemistry student claims: "It doesn't matter whether you use 22.71 or 24.8 L/mol as the molar volume — the difference is only about 10%, so it won't affect whether you pass or fail an exam question." Critically evaluate this claim with reference to a specific calculation. 4 MARKS
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Row 1 (2H₂O₂ → 2H₂O + O₂, STP):
MM(H₂O₂) = 34.015; n = 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 at STPRow 2 (CH₄ + 2O₂ → CO₂ + 2H₂O, RTP):
n(CO₂) = 12.4 ÷ 24.8 = 0.5000 mol; ratio CO₂:CH₄ = 1:1; n(CH₄) = 0.5000 mol MM(CH₄) = 16.043; m(CH₄) = 0.5000 × 16.043 = 8.02 gRow 3 (2KClO₃ → 2KCl + 3O₂, RTP):
MM(KClO₃) = 122.55; n = 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 at RTP1. B — 24.8 L/mol. Room temperature (25°C) = RTP. STP (0°C) uses 22.71 L/mol.
2. C — 1.12 L. n(CaCO₃) = 5.00÷100.09 = 0.04996 mol; ratio 1:1; n(CO₂) = 0.04996; V = 0.04996×22.71 = 1.12 L at STP.
3. A — 3.60 g. n(H₂) = 4.96÷24.8 = 0.2000 mol; ratio H₂:H₂O = 2:2 = 1:1; n(H₂O) = 0.2000; MM(H₂O) = 18.015; m = 0.2000×18.015 = 3.60 g.
4. D — 6.00 L. All gases at same T/P → volume ratio = coefficient ratio. H₂:NH₃ = 3:2; V(NH₃) = 9.00×(2÷3) = 6.00 L.
5. B. V = n × molar volume. Using 24.8 instead of 22.71 gives a larger multiplier → larger (incorrect) volume. Error = (24.8−22.71)÷22.71 × 100 = 10.7% too high.
6. C. n(O₂) at STP: V = n × 22.71; n = 5.60 ÷ 22.71 = 0.2500 mol. At RTP: V = 0.2500 × 24.8 = 6.20 L. The classmate's principle is correct (same moles) but the number 5.68 L is wrong — the correct RTP equivalent is 6.20 L.
7. A. The volume ratio shortcut (coefficient ratio = volume ratio) applies only when all reactants and products involved are gases at the same temperature and pressure. H₂ + Cl₂ → 2HCl satisfies this — all three species are gases. The other options involve solids, solutions, or liquids where the shortcut cannot be used.
Q6 (5 marks):
(a) m(CaCO₃) = 500,000 g; MM(CaCO₃) = 100.09 n = 500,000 ÷ 100.09 = 4996 mol; ratio 1:1; n(CO₂) = 4996 mol V(CO₂) = 4996 × 22.71 = 111,900 L ≈ 1.12 × 10⁵ L at STP(b) The actual volume would differ because: (i) the kiln operates at high temperature, not STP — gases expand at higher temperatures, so the actual volume would be much larger than calculated; (ii) the reaction may not proceed to 100% completion (yield < 100%), giving less CO₂ than theoretically predicted; (iii) the limestone may contain impurities, reducing the effective mass of CaCO₃ available.
Q7 (5 marks):
(a) n(Zn) = 6.54÷65.38 = 0.1000 mol; ratio 1:1; n(H₂) = 0.1000 mol V(H₂) at RTP = 0.1000 × 24.8 = 2.48 L (b) Student's answer using STP: V = 0.1000 × 22.71 = 2.24 L % error = (2.48 − 2.24) ÷ 2.48 × 100 = 9.68% ≈ 9.7%Q8 (3 marks): The volume ratio shortcut can be applied to H₂ + Cl₂ → 2HCl because all three species in this reaction are gases at the same temperature and pressure. By Avogadro's law, equal volumes of gases contain equal numbers of moles, so the volume ratio equals the coefficient ratio directly (1 L H₂ reacts with 1 L Cl₂ to give 2 L HCl). In contrast, CaCO₃ and CaO are solids — they do not occupy measurable gas volumes. The mole concept still applies to them, but their "volume" in the Avogadro's law sense is not relevant. To find the volume of CO₂ produced, you must first convert the mass of CaCO₃ to moles (using n = m ÷ MM), apply the mole ratio, then convert CO₂ moles to volume. The shortcut cannot be used for any reaction involving solids or liquids.
Q9 (4 marks):
(a) Student 1 (full method): MM(C₃H₈) = 3(12.011)+8(1.008) = 44.094; n = 44.1÷44.094 = 1.000 mol C₃H₈:CO₂ ratio = 1:3; n(CO₂) = 3.000 mol; V = 3.000 × 24.8 = 74.4 L Student 2 (volume ratio): all species gases → volume ratio = coefficient ratio V(C₃H₈) = n × 24.8 = 1.000 × 24.8 = 24.8 L; V(CO₂) = 24.8 × 3 = 74.4 L(b) Both methods give the same result (74.4 L) because all reactants and products are gases at the same conditions. The volume ratio shortcut is valid here because Avogadro's law applies to all gases equally. Note: the question values of 66.7 L and 66.5 L were illustrative — actual correct answer is 74.4 L at RTP.
Q10 (4 marks): The claim is incorrect. Choosing the wrong molar volume introduces a systematic 10.7% error in every gas volume calculation. In an exam, this means the final numerical answer is wrong, typically losing the answer mark even if working is shown. Example: n = 0.500 mol of CO₂; correct V(RTP) = 0.500 × 24.8 = 12.4 L; with wrong value: 0.500 × 22.71 = 11.2 L — off by 1.2 L. In a 3-mark calculation, a wrong numerical answer usually means losing at least the final mark. Furthermore, if the error propagates into a subsequent calculation (e.g., finding mass from the wrong volume), multiple steps are compromised. The student should always identify the stated conditions before choosing the molar volume.
Gas Stoichiometry
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