5.17C - Gas volumes and balanced equations

Molar volume toolkit

Molar volume is the volume occupied by one mole of any gas at room temperature and pressure.

Use the version that matches the unit you want:

For a solid, use mass and relative formula mass to find moles.

Here, n is amount in moles, m is mass in grams, and $M_{r}$ is relative formula mass. $M_{r}$ itself has no units, but its numerical value is the same as the molar mass in g mol-1 for one mole of the substance. If the mass is given in kilograms, convert it to grams before using the formula:

1 kg = 1000 g

For a gas at room temperature and pressure:

So, in $\mathrm{dm}^{3}$:

volume of gas = n x 24

And in $\mathrm{cm}^{3}$:

volume of gas = n x 24000

The balanced equation supplies the mole ratio. The ratio is what connects the solid to the gas; do not assume the same number of moles unless the equation shows a 1:1 ratio.

Solid mass to gas volume

Use this route when the question gives the mass of a solid and asks for the volume of a gas.

1. Write the balanced equation ratio between the solid and the gas.
2. Calculate moles of the solid from mass / Mr.
3. Use the ratio to calculate moles of the gas.
4. Use the molar volume to calculate the gas volume.
5. Convert units if needed and give the answer with units.

Worked example 1

Calcium carbonate decomposes when heated.

CaCO3(s) -> CaO(s) + CO2(g)

Calculate the volume of carbon dioxide, in $\mathrm{cm}^{3}$, formed from 5.00 g of calcium carbonate at room temperature and pressure.
Relative atomic masses: $\ce{Ca} = 40.0$, $C = 12.0$, $O = 16.0$.
Molar volume at room temperature and pressure = $24\,\mathrm{dm}^{3}\,\mathrm{mol}^{-1}$.

Balanced equation ratio:
1 mol CaCO3 : 1 mol CO2

Mr of CaCO3:
40.0 + 12.0 + (3 x 16.0) = 100.0

Moles of CaCO3:
n = mass / Mr
n = 5.00 / 100.0
n = 0.0500 mol

Moles of CO2 from the ratio:
1 mol CaCO3 gives 1 mol CO2
n(CO2) = 0.0500 mol

Use molar volume:
volume of CO2 = n x 24
volume of CO2 = 0.0500 x 24
volume of CO2 = 1.20 dm3

Unit conversion:
1.20 dm3 x 1000 = 1200 cm3

The volume of carbon dioxide is $1200\,\mathrm{cm}^{3}$.

Using non-1:1 ratios

Many Edexcel mistakes happen at the ratio step. The coefficients in the balanced equation are mole ratios. If the equation says 2 mol of one substance produce 1 mol of a gas, the gas amount is half the amount of that substance.

Worked example 2

Sodium nitrate decomposes when heated.

2NaNO3(s) -> 2NaNO2(s) + O2(g)

Calculate the volume of oxygen, in $\mathrm{dm}^{3}$, formed from 17.0 g of sodium nitrate at room temperature and pressure.
Relative atomic masses: $\ce{Na} = 23.0$, $N = 14.0$, $O = 16.0$.
Molar volume at room temperature and pressure = $24\,\mathrm{dm}^{3}\,\mathrm{mol}^{-1}$.

Balanced equation ratio:
2 mol NaNO3 : 1 mol O2

Mr of NaNO3:
23.0 + 14.0 + (3 x 16.0) = 85.0

Moles of NaNO3:
n = mass / Mr
n = 17.0 / 85.0
n = 0.200 mol

Moles of O2 from the ratio:
2 mol NaNO3 gives 1 mol O2
n(O2) = 0.200 x (1 / 2)
n(O2) = 0.100 mol

Use molar volume:
volume of O2 = n x 24
volume of O2 = 0.100 x 24
volume of O2 = 2.40 dm3

Unit conversion check:
2.40 dm3 = 2400 cm3

The volume of oxygen is $2.40\,\mathrm{dm}^{3}$.

The ratio step can also increase the number of gas moles. For $\ce{2Al (s) + 3Cl2 (g) -> 2AlCl3 (s)}$, the ratio is:

2 mol Al : 3 mol Cl2