If ten volume of dihydrogen gas react with five volume of dioxygen gas, how many volume of water vapour would be produced?
2${ H }_{ 2 }$ +
${ O }_{ 2 }$ ---------------->
2 ${ H }{ 2 }O$
According to Gay Lussac’s law of gaseous volumes, 2 volume of dihydrogen react with 1 volume of ${ O }{ 2 }$ to produce 2 volume of water vapour. Therefore, 10 volume of dihydrogen on reaction with 5 volume of dioxygen will produce 10 volume of water vapour.