What volume of air at N.T.P containing 21 % oxygen by volume is required to completely burn 1000 g of sulphur containing 4 % incombustible matter ?

Explanation

The balanced chemical equation is:
`underset(32 g) underset( 1 mol)(S + O_(2)) to underset(22.4 L)underset(1 "mol")(SO_(2))`
Amount of sulphur in the sample
`=(1000 xx 96)/100 = 960 g`
32 g of sulphur require oxygen at N.T.P. = 22.4 L
960 g of sulphur require oxygen at N.T.P.
`=(22.4)/32 xx 960`
=672 L
Volume of air required `=(672 xx 100)/21= 3200 L`