If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Explanation

`underset("2 volume")(2H_(2)(g))+underset("1 volume")(O_(2)(g)) to underset("2 volume")(2H_(2)O(g))`
According to Gay Lussac's law of gaseous volumes, 2 volume of hydrogen reacts with 1 volume of oxygen to produce 2 volume of water vapour under the same conditions of temperature and pressure. therefore, 10 volume of hydrogen reacts with 5 volume of oxygen to produce 10 volume of water vapour at the same temperature and pressure.