`H_(2)` and `O_(2)` react according to the equation
`2H_(2)(g)+O_(2)rarr 2H_(2)O(g)`
Thus, 2 volumes of `H_(2)` react with `1` volume of `O_(2)` to produce `2` volume of water vapour. Hence, `10` volumes of `H_(2)` will react completely with `5` volumes of `O_(2)` to produce `10` volumes of water vapour