Find the moles of gas from the mole ratio, then multiply by the molar volume: V = n x 24 dm3 mol−1 at room conditions (or 22.4 dm3 mol−1 at STP).
This page walks through volume of gas in a reaction for SPM Chemistry, step by step: the formula you need, the units to watch, and a worked example.
When you use this
Many SPM Chemistry reactions give off a gas, hydrogen from a metal and acid, carbon dioxide from a carbonate and acid. When the question asks for the volume of that gas, this is the calculation.
The formula and units
V = n x molar volume, where n is the moles of gas.
- At room conditions the molar volume is 24 dm3 mol−1.
- At standard temperature and pressure (STP) it is 22.4 dm3 mol−1.
Rearranged, n = V / molar volume.
Units. Volume V in cubic decimetres (dm3), amount n in mol, molar volume in dm3 mol−1. If a question uses cm3, remember 1 dm3 = 1000 cm3. Always read whether the question is at room conditions or at STP and use the matching molar volume.
Worked example 1 (easy)
How many moles of gas are present in 12 dm3 of a gas measured at room conditions?
- n = V / molar volume = 12 / 24 = 0.5 mol.
Worked example 2 (medium)
Zinc reacts with dilute hydrochloric acid: Zn + 2HCl → ZnCl2 + H2. Find the volume of hydrogen produced at room conditions from 0.20 mol of zinc.
- Ratio Zn : H2 = 1 : 1, so n(H2) = 0.20 mol.
- V(H2) = 0.20 x 24 = 4.8 dm3.
Worked example 3 (SPM level)
Calcium carbonate reacts with excess dilute hydrochloric acid: CaCO3 + 2HCl → CaCl2 + H2O + CO2. Find the volume of carbon dioxide released at room conditions from 10 g of calcium carbonate. (Ca = 40, C = 12, O = 16)
- M(CaCO3) = 40 + 12 + (16 x 3) = 100 g mol−1.
- n(CaCO3) = 10 / 100 = 0.10 mol.
- Ratio CaCO3 : CO2 = 1 : 1, so n(CO2) = 0.10 mol.
- V(CO2) = 0.10 x 24 = 2.4 dm3 (or 2400 cm3).
If the same question were set at STP, you would use 22.4 dm3 mol−1 instead: V = 0.10 x 22.4 = 2.24 dm3.
Avogadro’s law behind it
Why can such different gases share one molar volume? Because equal volumes of gases at the same temperature and pressure contain equal numbers of particles. One mole of any gas therefore occupies the same volume, 24 dm3 at room conditions, 22.4 dm3 at STP, whether it is light hydrogen or heavy carbon dioxide. That single idea is what lets you jump straight from moles to volume in one multiplication. It also explains the boundaries of the method: it applies only to the gases in the equation, never to a solid such as calcium carbonate or a liquid such as water, so you must first find the moles of gas from the mole ratio and only then reach for the molar volume.
Common traps
- Using the wrong molar volume. 24 dm3 mol−1 is for room conditions; at STP use 22.4 dm3 mol−1. Read the question.
- cm3 and dm3 mixed up. Convert with 1 dm3 = 1000 cm3 before or after, but stay consistent.
- Skipping the mole ratio. Find the moles of gas from the equation first; do not multiply the moles of solid by the molar volume.
- Assuming the solid has a molar volume. Only gases do; the molar volume applies to CO2 or H2, not to CaCO3 or Zn.
In our online lessons we make students state the conditions before touching the numbers, so the right molar volume is chosen every time, a small habit that protects easy marks in Paper 2 structured questions.
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