Gas Stoichiometry
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.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
Gas Stoichiometry Formulas
V = n × molar volume (moles → gas volume)
STP (0°C, 100 kPa): molar volume = 22.71 L/mol
RTP (25°C, 100 kPa): molar volume = 24.8 L/mol
Key facts
- STP = 0°C, 100 kPa → 22.71 L/mol (NESA standard)
- RTP = 25°C, 100 kPa → 24.8 L/mol
- n = V ÷ molar volume (volume to moles)
- V = n × molar volume (moles to volume)
Concepts
- Convert gas volume to moles and vice versa
- Find gas volume produced from a solid reactant mass
- Find mass of reactant needed to produce a given gas volume
- Apply gas steps to both inputs and outputs of reactions
Skills
- TODO
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.
STP — Standard Temperature and Pressure
- Temperature: 0°C (273 K)
- Pressure: 100 kPa
- Used when question says "STP" or "0°C, 100 kPa"
- Current NESA standard. Note: 22.4 L/mol is the older value at 0°C and 1 atm (101.325 kPa) — you may see it in older resources. NESA uses 22.71 L/mol at 0°C and 100 kPa.
RTP — Room Temperature and Pressure
- Temperature: 25°C (298 K)
- Pressure: 100 kPa
- Used when question says "RTP", "room conditions", or "25°C"
- More realistic for laboratory experiments
If a gas volume is the input (given) → use Step 0: n = V ÷ MV, then proceed to Step 3 directly.
If a gas volume is the output (asked for) → complete Steps 1–3 normally, then use Step 5: V = n × MV.
If both input and output are gases → use Step 0 AND Step 5.
Avogadro's law: equal volumes of all gases at the same T and P contain equal numbers of molecules. Molar volume: STP (0 °C, 100 kPa) = 22.71 L mol⁻¹; RTP (25 °C, 100 kPa) = 24.8 L mol⁻¹. To find moles from gas volume: n = V ÷ Vm. Always identify conditions before choosing a Vm value.
Pause — copy the highlighted rule and values into your book before moving on.
Odd one out — three of these conditions all use 24.8 L/mol as the molar volume. Which one doesn't belong?
Worked examples · reveal as you go
What volume of CO₂ is produced at STP when 25.0 g of CaCO₃ decomposes? CaCO₃ → CaO + CO₂. (Ca=40.078, C=12.011, O=15.999)
What volume of O₂ at RTP is required to completely burn 0.500 mol of C₂H₆? 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.
What mass of Zn is needed to produce 3.72 L of H₂ at RTP? Zn + 2HCl → ZnCl₂ + H₂. (Zn = 65.38)
In 2H₂ + O₂ → 2H₂O (gas), what volume of H₂O vapour forms from 4.00 L of H₂ at constant temperature and pressure?
Click two steps to swap them. Order the gas-stoichiometry method to solve: what volume of CO₂ is produced at STP when 25.0 g of CaCO₃ decomposes? (CaCO₃ → CaO + CO₂)
- Apply the mole ratio CaCO₃ : CO₂ = 1 : 1, so n(CO₂) = 0.2498 mol.
- Identify the conditions (STP → use 22.71 L/mol) and confirm the equation is balanced.
- Final answer: V(CO₂) = 5.67 L at STP.
- Convert mass to moles: MM(CaCO₃) = 100.09; n(CaCO₃) = 25.0 ÷ 100.09 = 0.2498 mol.
- Convert moles of gas to volume: V = n × Vₘ = 0.2498 × 22.71.
Common errors · the 3 traps that cost marks
Using 22.71 L/mol for RTP conditions (or 24.8 for STP)
This is the single most tested trap in gas stoichiometry. The question will almost always specify conditions — read for "STP", "standard conditions", "0°C" (use 22.71), or "RTP", "room temperature", "25°C" (use 24.8). Using the wrong value gives an answer that is off by a factor of 24.8 ÷ 22.71 = 1.107 — a 10.7% error that will cost marks even if all other steps are correct.
✓ Fix: Before any calculation, underline the conditions stated in the question. Write "STP → 22.71" or "RTP → 24.8" at the top of your working before you start.
Forgetting to convert mass to moles before applying the mole ratio
When a solid reactant mass is given and a gas volume is asked for, students sometimes skip Step 2 (n = m ÷ MM) and go straight from mass to volume using the molar volume. This is wrong — molar volume converts moles to litres, not grams to litres. You must convert mass → moles first, then apply the ratio, then convert moles → volume.
✓ Fix: Always go mass → moles → ratio → moles of gas → volume. Never skip the mass-to-moles step, even when the question asks for a gas volume.
Applying the gas volume ratio shortcut when reactants are not all gases
The volume ratio shortcut (volume ratio = coefficient ratio) only applies when ALL species in the comparison are gases at the same temperature and pressure. In CaCO₃ → CaO + CO₂, the CaCO₃ and CaO are solids — only CO₂ is a gas. You cannot say "1 L of CaCO₃ produces 1 L of CO₂" — solids don't have volumes in this sense. The shortcut works only for reactions like H₂ + Cl₂ → HCl, where all species are gases.
✓ Fix: Use the volume ratio shortcut only when every reactant and product you're comparing is explicitly a gas in the question. If any solid or liquid is involved, use the full 4-step method.
Quick-fire practice · 5 reps +2 XP per reveal
What volume does 2.0 mol of an ideal gas occupy at 25 °C and 100 kPa? (Vₘ = 24.79 L mol⁻¹)
How many moles of gas are in 12.4 L at 25 °C and 100 kPa? (Vₘ = 24.79 L mol⁻¹)
For N₂ + 3H₂ → 2NH₃, what volume of H₂ reacts with 10 L of N₂ (same temperature and pressure)?
What volume of CO₂ (25 °C, 100 kPa) forms when 0.20 mol of CaCO₃ decomposes? (CaCO₃ → CaO + CO₂)
Find the mass of 6.0 L of O₂ at 25 °C and 100 kPa. (Vₘ = 24.79 L mol⁻¹, M(O₂) = 32.00)
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.
Reflect: how did your initial thinking compare to what you've learned?
Write a reflection in your workbook.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Complete the gas-stoichiometry working below. Reaction: Zn + 2HCl → ZnCl₂ + H₂. What mass of Zn is needed to produce 3.72 L of H₂ at RTP? (Zn = 65.38)
Step 2 — Gas volume → moles: n(H₂) = V ÷ Vₘ = 3.72 ÷ = mol.
Step 3 — Mole ratio Zn : H₂ = : 1, so n(Zn) = 0.150 mol.
Step 4 — Moles → mass: m(Zn) = n × MM = 0.150 × 65.38 = g.
Q1. 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)
Q2. 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)
Q3. 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.
Q4. 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)
Q5. 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.
📖 Comprehensive answers (click to reveal)
Activity 2 — Data Table Answers
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 RTP❓ Multiple Choice
1. 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 = 1.135 ≈ 1.14 L at STP (using NESA standard 22.71 L/mol). Option C (1.12 L) is the closest provided answer and would be obtained using the older 22.4 L/mol value.
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 = 9.2% too high.
6. C. n(O₂) at STP: V = n × 22.71; n = 5.60 ÷ 22.71 = 0.2466 mol. At RTP: V = 0.2466 × 24.8 = 6.12 L. The classmate's principle is correct (same moles) but the number 5.68 L is wrong — the correct RTP equivalent is 6.12 L (using NESA 22.71 L/mol).
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.
Short Answer Model Answers
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 = 113,459 L ≈ 1.13 × 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 = 2.27 L % error = (2.48 − 2.27) ÷ 2.48 × 100 = 8.47% ≈ 8.5%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.
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