The big idea: Avogadro's law: at the same temperature and pressure, equal volumes of any gases contain equal numbers of moles.
So for reacting gases you can skip the moles entirely — the volume ratio is the same as the mole (coefficient) ratio from the balanced equation.
Example: in N2 + 3H2 → 2NH3, 1 volume of nitrogen reacts with 3 volumes of hydrogen to give 2 volumes of ammonia.
Two tools, two jobs: - Comparing gases at the same T and P? Use the volume ratio = coefficient ratio directly — no need for moles or molar volume.
- Converting between an amount and a volume? Use the molar volume: Vm = 22.7 dm³ mol⁻¹ at STP (273 K, 100 kPa).
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
Because volume is proportional to the amount of gas at fixed T and P, the coefficients in a balanced equation are also the volume ratio of the gases. Just scale up or down.
| Reaction (gases only) | Coefficient ratio | Volume ratio at same T, P |
|---|---|---|
| N2 + 3H2 → 2NH3 | 1 : 3 : 2 | 1 vol N2 + 3 vol H2 → 2 vol NH3 |
| 2CO + O2 → 2CO2 | 2 : 1 : 2 | 2 vol CO + 1 vol O2 → 2 vol CO2 |
| C2H4 + 3O2 → 2CO2 + 2H2O | 1 : 3 : 2 : 2 | for the gases, 1 : 3 : 2 (water may condense) |
Worked example — volume of oxygen needed
What volume of oxygen is needed to completely burn 25 cm³ of methane, CH4, all volumes measured at the same temperature and pressure?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Solution
- Read the ratio first — from the equation, CH4 : O2 is 1 : 2:
- Volume ratio = coefficient ratio, so multiply the methane volume by 2:
- Work it out — keep the unit:
Final answer
50 cm³ of oxygen (the volume ratio is just the 1 : 2 mole ratio).
Memorize terms 3x faster
Smart flashcards show you cards right before you forget them. Perfect for definitions and key concepts.
When a question links a mass or amount to a gas volume, switch to the molar volume. At STP one mole of any gas occupies 22.7 dm³, so divide a volume by 22.7 to get moles, or multiply moles by 22.7 to get a volume.
- amount of gas (mol)
- volume of the gas (dm³)
- molar volume = 22.7 dm³ mol⁻¹ at STP (273 K, 100 kPa)
Watch the units: Molar volume is 22.7 dm³ mol⁻¹ — so volumes must be in dm³ here.
Convert first if needed: 1 dm³ = 1000 cm³, so divide a cm³ value by 1000 before using 22.7.
Worked example — volume of gas produced
0.050 mol of calcium carbonate reacts completely with excess hydrochloric acid. Calculate the volume of carbon dioxide produced, measured at STP.
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Solution
- The mole ratio CaCO3 : CO2 is 1 : 1, so the amount of CO2 equals the amount of CaCO3:
- Formula first — volume from amount at STP:
- Substitute the molar volume 22.7 dm³ mol⁻¹:
- Work it out — keep the unit:
Final answer
V = 1.1 dm³ of CO₂ at STP.
Worked example — amount of gas from a volume
A reaction releases 454 cm³ of hydrogen gas, measured at STP. Calculate the amount, in moles, of hydrogen produced.
Solution
- Convert the volume to dm³ first (1 dm³ = 1000 cm³):
- Formula first — amount from volume at STP:
- Substitute:
- Work it out:
Final answer
n = 0.0200 mol of hydrogen.
How this is tested: Reacting gas volumes is a favourite Paper 1A MCQ and a short Paper 2 part.
- Paper 1A: a one-step volume-ratio question — 'what volume of CO2 is made from this volume of fuel?' — or finding the unreacted excess gas left over. - Paper 2: 'find the maximum product volume' and then the total gas volume in the container after the reaction, all at the same T and P.
The recurring trap: forgetting to subtract the gas that reacted when asked for the volume remaining, and counting liquid water as a gas volume.
Score it cleanly: (1) Balance the equation and read the gas volume ratio. (2) Same T and P → work directly in volumes, no moles needed. (3) For 'remaining' or 'total', track each gas: start − reacted + made, and ignore any species that is not a gas.
IB-style question — combustion of propane (a)
20 cm³ of propane, C3H8, is mixed with 120 cm³ of oxygen and ignited; the propane burns completely. All volumes are measured at the same temperature and pressure (with water as a liquid).
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
(a) Determine the volume of oxygen that remains unreacted. [2]
How to score the marks
- Mark 1 — oxygen used. The ratio C3H8 : O2 is 1 : 5, so 20 cm³ of propane uses 5 × 20:
- Mark 2 — oxygen left. Subtract what reacted from what was supplied:
Final answer
20 cm³ of oxygen is left unreacted.
IB-style question — combustion of propane (b)
(b) Determine the total volume of gas in the container after the reaction, measured at the same temperature and pressure. [2]
How to score the marks
- Mark 1 — gases present after reaction. The water is a liquid, so the only gases are the CO2 produced and the leftover O2. CO2 made = ratio 1 : 3, so 3 × 20:
- Mark 2 — add the gas volumes (60 cm³ CO2 + 20 cm³ unreacted O2); the liquid water adds nothing:
Final answer
80 cm³ of gas in total (60 cm³ CO₂ + 20 cm³ unreacted O₂; the water is liquid).