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Gas volumes and reacting amounts

Convert gas volumes into moles at room temperature and pressure, then use a balanced equation to calculate reacting masses and gas volumes.

  • SEC G3 Pure Chemistry 2027

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Gas volume can tell you how much gas you have, provided you know its temperature and pressure. A balanced equation then connects that amount to the amount of another reactant or product.

Be comfortable with moles and molar mass and balanced equations. Solution volumes use a different method, taught in Solution concentration and titration calculations.

Molar volume depends on temperature and pressure

Molar volume, Vₘ, is the volume occupied by one mole of gas at a specified temperature and pressure. In this course’s calculations, use:

Vₘ = 24 dm³ mol⁻¹ = 24 000 cm³ mol⁻¹ at room temperature and pressure (RTP)

This is the syllabus approximation. At the same temperature and pressure, gases are treated as having the same molar volume. Do not use the RTP value for a gas at other conditions unless the question tells you to use it; calculations with gas laws are not required here.

For a gas amount n and volume V:

n = V/Vₘ V = nVₘ

Keep volume units compatible. Use dm³ with 24 dm³ mol⁻¹, or cm³ with 24 000 cm³ mol⁻¹. Remember that 1000 cm³ = 1 dm³.

Gas volume and amount at room temperature and pressureGas volume V in cubic decimetres converts to amount n in moles by dividing by molar volume, 24 cubic decimetres per mole at RTP. Reverse by multiplying by molar volume. The two arrows point in opposite directions.Gas measured at RTPGas volume, Vdm³÷ Vₘ× VₘRTPAmount, nmoln = V/VₘV = nVₘVₘ = 24 dm³ mol⁻¹
At RTP, divide a gas volume in dm³ by 24 dm³ mol⁻¹ to find its amount. Reverse the conversion by multiplying. The same calculation can use cm³ with 24,000 cm³ mol⁻¹; keep the volume units compatible.

Use the equation between the conversions

Given amount → mole ratio → required amount
  1. Convert the given quantity to moles: use n = m/M for a mass, or n = V/Vₘ for a gas volume.
  2. Use the coefficients in the balanced equation to find the required substance’s amount.
  3. Convert that amount to the requested mass or gas volume.

If all the volumes compared are gases at the same temperature and pressure, their volume ratio equals their mole ratio. For example, in 2CO(g) + O₂(g) → 2CO₂(g), two volumes of carbon monoxide react with one volume of oxygen to form two volumes of carbon dioxide, measured at the same conditions. This shortcut does not apply to a solid, liquid or aqueous solution.

When a reactant is in excess, the other reactant controls the theoretical product amount. If both reactant quantities are given, use Limiting reactants to check which one runs out first.

Set the volumes of hydrogen and oxygen in the syringes, then spark the mixture and see how much gas is left.

t = 0.00 s

60 cm³ of hydrogen and 50 cm³ of oxygen at room temperature and pressure. Amounts now: H₂ 0.00250 mol, O₂ 0.00208 mol, H₂O 0 mol.

n(H2)
0.00250 mol
n(O2)
0.00208 mol
Gas left
— cm³
Limiting reactant
—
Reaction
cm³
cm³

Try this

0 of 4 done

  1. Choose magnesium and acid that react with nothing left over, then start. (not done yet)

  2. Run Mg + HCl once with the magnesium used up and once with the acid used up. (not done yet)

  3. Mix hydrogen and oxygen so that no gas is left after the spark. (not done yet)

  4. Weigh the crucible part-way through heating, then again once its mass stops changing. (not done yet)

Worked examples

Modelled example 1

From gas volume to amount

Core

Problem

Calculate the amount, in mol, of 600 cm³ of oxygen gas at RTP. Give your answer to three significant figures.

Study the worked solution

  1. Convert the gas volume

    Method

    Divide the volume in cm³ by 1000.

    Reason

    The RTP molar volume is expressed in dm³ mol⁻¹.

    Working

    V = 600/1000 = 0.600 dm³.

  2. Use molar volume

    Method

    Divide the gas volume by 24 dm³ mol⁻¹.

    Reason

    Use the syllabus molar volume of 24 dm³ mol⁻¹ at RTP.

    Working

    n = V/24 = 0.600/24 = 0.0250 mol.

  3. Check the scale

    Method

    Compare 0.600 dm³ with 24 dm³.

    Reason

    The sample is 1/40 of the molar volume, so it should contain 1/40 mol.

    Working

    n(O₂) = 0.0250 mol.

Guided practice 2

From a reacting mass to a gas volume

About 7 min

Problem

5.00 g of pure calcium carbonate decomposes completely on heating:

CaCO₃(s) → CaO(s) + CO₂(g)

Find the theoretical volume of carbon dioxide at RTP, using M(CaCO₃) = 100 g mol⁻¹ and the syllabus molar volume. Assume all the gas is collected and measured at RTP.

Convert to moles before using the equation

Hints

Hint 1: start from the solid mass

Use n = m/M for calcium carbonate; molar gas volume does not apply to this solid.

Hint 2: read the ratio

The coefficients give CaCO₃:CO₂ = 1:1.

View solution step by step

  1. Find the amount of the solid

    Method

    Divide its mass by its molar mass.

    Reason

    The equation relates amounts in mol, rather than equal masses.

    Working

    n(CaCO₃) = 5.00/100 = 0.0500 mol.

  2. Use the balanced equation

    Method

    Apply the 1:1 mole ratio.

    Reason

    Complete decomposition gives one mole of carbon dioxide per mole of calcium carbonate.

    Working

    n(CO₂) = 0.0500 mol.

  3. Find the gas volume

    Method

    Multiply the gas amount by its molar volume at RTP.

    Reason

    The gas is measured at RTP after collection.

    Working

    V(CO₂) = 0.0500(24) = 1.20 dm³.

Try it yourself

Mind stretcher 1: Use the equation to find a gas volumeExtension

Question: Magnesium reacts with dilute hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). Assuming the acid is in excess and all the magnesium reacts, what volume of H₂(g) at RTP is produced from 0.050 mol of Mg?

Show answer

Mole ratio Mg:H₂ is 1:1, so n(H₂) = 0.050 mol.

At RTP, V = n × 24:

V = 0.050 × 24 = 1.2 dm³

Mind stretcher 2: Use a gas volume ratio directlyExtension

Carbon monoxide burns in oxygen: 2CO(g) + O₂(g) → 2CO₂(g). Find the minimum oxygen volume needed to burn 120 cm³ of carbon monoxide completely, and the carbon dioxide volume produced. All gas volumes are measured at the same temperature and pressure.

Show answer

The gas volume ratio is CO:O₂:CO₂ = 2:1:2. Oxygen required is 120/2 = 60 cm³; carbon dioxide produced is 120 cm³. Using mole ratios directly saves a conversion because all three quantities are gas volumes at the same conditions.

Solution concentration uses solute amount and final solution volume. These related explanations cover the methods:

Practise and check

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