What volume of air containing `21%` of oxygen by volume is required to completely burn `1 Kg` of sulphur `(S_(8))` which contains `4%` incombustible material? Sulphur burns according to the reaction
`(1)/(2)S_(8)+O_(2) rarr SO_(2)`

Explanation

The amountof incombutstible material is `1000xx(4)/(100)=40 g`
So the mass of pure sulphur is `1000 g-40 g=960 g`
Calculation of amount of pure `O_(2)` required:
`(1)/(8)S_(8)+O_(2) to SO_(2)`
From the chemical equation, it is evident that `32 g` of sulphur requires `22.4 L` of oxygen at `STP`.
Therefore, `960 g` of sulphur require oxygen
`=(22.4xx960 g)/(32 g)=672 L`
Calculation of volume of air:
If `21 L` of `O_(2)` is needed, air required is `100 L`.
If `672 L` of `O_(2)` is needed, air required is
`(100xx672 L)/(21 L)=3200 L`