When a reacting species is a gas, pressure, volume and temperature can be converted into moles using the ideal-gas law. Once the amount in moles is known, ordinary reaction stoichiometry applies.

The general calculation path is

$$p,V,T\longrightarrow n=\frac{pV}{R_uT}\longrightarrow\text{stoichiometric ratio}\longrightarrow\text{desired amount or gas volume}.$$

Example

Hydrogen reacts with oxygen according to

$$\mathrm{2H_2(g)+O_2(g)\rightarrow2H_2O}.$$

Suppose $5.00,\mathrm L$ of oxygen is available at $100,\mathrm{kPa}$ and $300,\mathrm K$, with excess hydrogen.

First find the oxygen amount:

$$n_{O_2}=\frac{pV}{R_uT} =\frac{(100,\mathrm{kPa})(5.00,\mathrm L)}{(8.314,\mathrm{kPa,L,mol^{-1},K^{-1}})(300,\mathrm K)} \approx0.200,\mathrm{mol}.$$

The balanced equation produces two moles of water per mole of oxygen, so

$$n_{H_2O}=0.200\times\frac{2}{1}=0.401,\mathrm{mol}.$$

If a gaseous product volume is requested at specified conditions, convert the stoichiometric mole amount back using

$$V=\frac{nR_uT}{p}.$$

Equal-temperature, equal-pressure shortcut

For ideal gases compared at the same temperature and pressure, volume is proportional to moles. Gas-volume ratios then equal stoichiometric mole ratios. For

$$\mathrm{N_2+3H_2\rightarrow2NH_3},$$

one volume of nitrogen reacts with three equal-condition volumes of hydrogen to form two equal-condition volumes of gaseous ammonia, provided all gases are compared at the same $T$ and $p$.

This shortcut is Avogadro's law applied to stoichiometry; it is not valid when the gas volumes refer to different temperatures or pressures.

Gas measurements therefore change how amounts are obtained or reported, not the stoichiometric relationships encoded by the balanced chemical equation.