5.18C - Gas volume calculations using Avogadro's law

The Avogadro's Law Idea

Avogadro's law says that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. For calculations, that means the volume of a gas is directly proportional to the amount of gas in moles, as long as temperature and pressure stay the same.

The important GCSE consequence is not that all gases have the same mass. Hydrogen, oxygen and carbon dioxide have different masses. The useful point is that if the gases are measured under the same conditions, mole ratios in a balanced equation can be used as volume ratios.

For example, this balanced equation shows one molecule of hydrogen reacts with one molecule of chlorine to make two molecules of hydrogen chloride:

$\ce{H2 (g) + Cl2 (g) -> 2HCl (g)}$

So the gas volume ratio is:

$\ce{H2} : \ce{Cl2} : \ce{HCl} = 1 : 1 : 2$

This means 10 cm3 of hydrogen reacts with 10 cm3 of chlorine to make 20 cm3 of hydrogen chloride, if all the gases are measured at the same temperature and pressure.

From Equation to Volume Ratio

Only use the numbers in front of the formulae in the balanced equation. These are the coefficients. If there is no number in front of a formula, the coefficient is 1.

Do not use the small numbers inside formulae as the reacting ratio. In $\ce{CO2}$, the small 2 tells you that one carbon dioxide molecule contains two oxygen atoms. It does not tell you that carbon dioxide reacts in a volume ratio of 2.

Use this route:

1. Check the equation is balanced.
2. Write the coefficient ratio for the gases named in the question.
3. Use the known gas volume to scale the ratio.
4. Give the answer in the same unit, unless the question asks for a conversion. Do not compare a gas volume with a liquid or solid volume: for example, water vapour in an equation is H2O(g), whereas condensed H2O(l) has a very different volume.

Some examples:

The ratios describe volumes reacting or theoretically formed, not the entire initial mixture when a reactant is in excess.

This formula is just the balanced-equation ratio written as a calculation. It works here because the substances being compared are gases measured under the same conditions.

Worked Volume Calculations

Worked example 1:

Carbon monoxide reacts with oxygen:

$\ce{2CO (g) + O2 (g) -> 2CO2 (g)}$

Calculate the volume of oxygen needed to react completely with 80 cm3 of carbon monoxide. All gas volumes are measured at the same temperature and pressure.

The coefficient ratio is:

$\ce{CO} : \ce{O2} = 2 : 1$

So oxygen volume is half the carbon monoxide volume:

$80\,\mathrm{cm}^{3} \times 1/2 = 40\,\mathrm{cm}^{3}$

Answer: $40\,\mathrm{cm}^{3}$ of oxygen.

Notice the unit stayed as $\mathrm{cm}^{3}$ all the way through. There is no need to convert to $\mathrm{dm}^{3}$ unless the question asks for the answer in $\mathrm{dm}^{3}$.