This concept of reacting volumes may be easily applied to calculate excess and limiting reagents in gaseous reactions. 
Limiting gas volume
By inspection of the volumes reacting, it is a simple matter to determine the limiting reagent and which is in excess.
N_{2}

+

3H_{2}

2NH_{3}


1 volume


3 volumes


2 volumes

Experimental data



300 cm^{3}


300 cm^{3}


???

excess


limiting reagent

200cm^{3} ammonia

In the table above, the amounts of nitrogen and hydrogen provided initially are 300 cm^{3} of each.
But from the stoichoimetry of the equation we can see that 300cm^{3} of nitrogen gas would require 3 x that volume of hydrogen gas for complete reaction. The nitrogen is therefore in excess and only 100cm^{3} of the nitrogen can react with hydrogen.
The reaction proportions then are 100 cm^{3} nitrogen reacting with 300 cm^{3} hydrogen to produce 200 cm^{3} of ammonia.
Note: If only 100cm^{3} nitrogen reacts there is also 200 cm^{3} nitrogen left over at the end of the reaction.
Example: Calculate (i) the volume of carbon dioxide produced when 300 cm^{3} of methane burns in 1000 cm^{3} oxygen and (ii) the final volume of the reaction mixture. The equation for the reaction: CH_{4}(g) + 2O_{2}(g) CO_{2}(g) + 2H_{2}O(l) The ratio of gas volumes is 1 volume of methane reacts with 2 volumes of oxygen producing 1 volume of carbon dioxide (and liquid water, no volume) There is 300 cm^{3} methane  this needs 2 x 300 cm^{3} = 600 cm^{3} oxygen to burn completely and produce 1 x 300 cm^{3} carbon dioxide = 300cm^{3} carbon dioxide During the course of the reaction 300 cm^{3} methane and 600 cm^{3} of oxygen is used up from an initial total of 300 cm^{3} of methane and 1000 cm^{3} oxygen. There is 400 cm^{3} of oxygen in excess therefore the final mixture contains 300 cm^{3} carbon dioxide + 400 cm^{3} of oxygen = 700 cm^{3} of gas 