Gay-Lussac’s Law of Combining Volume states that when gases react, they do so in volumes, which are simple ratios to one another and to the volume of their product if gaseous, provided the temperature and pressure remain constant.

**Example**: 40 cm^{3} of carbon (II) oxide are sparked with 60cm^{3} of oxygen. If all the volumes of gases are measured at S.T. P, calculate the volume of excess reactant and name the gas.

**Solution:**

2CO_{(g)} + O_{2(g)} 2CO_{2(g)}

2 : 1

Since the CO and O_{2} are in 2:1 respectively.

2CO_{(g)} + O_{2(g)} 2CO_{2(g)}

40cm_{3} : 20cm_{3}

So, the reactant in excess is oxygen. The volume of excess oxygen = 60 – 20 = 40cm^{3}