Calculate the volume of oxygen required to burn completely a mixture of 22.4 `dm^(3)` of `CH_(4)` and 11.2 `dm^(3)` of `H_(2)` (all volumes measured at STP) `[1 dm^(3) = "litre"]`.

Explanation

(i) `underset(1 vol)(CH_(4)) + underset(2 vol) (2O_(2)) to underset(1 vol + 2 vol) (CO_(2) + 2H_(2)O)`,
(ii) `underset(2 vol)(2H_(2)) + underset(1 vol)(O_(2)) to underset(2 vol)(2H_(2)O)`
Total oxygen required = 44.8 `dm^(3) + 5.6 dm^(3) = 50.4 dm^(3)` [lits].
`therefore` the number of moles in 7g of nitrogen is 0.25 moles.