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

Mass of sulphur = 1000 g
Mass of combustible matter `= (1000 xx 4)/(100)=40` g
Mass of combustible matter (pure sulphur) `= 1000 - 40 = 960` g
Step I. Volume of oxygen required
The equation for the combustion reaction is :
`underset(32g)(S)+underset(22.4L)(O_(2))rarrSO_(2)`
32g of sulphur require `O_(2)=22.4 L`
1 g of sulphur require `O_(2)=(22.4)/(32)L`
960 g of sulphur require `O_(2)=(22.4)/(32)xx960 = 672 L`
Step II. Volume of air required
If `O_(2)` is 21 L, then air `= 100 L`
If `O_(2)` is 1 L, then air `= (100)/(21)L`
If `O_(2)` is 672 L, then air `= (100)/(21) xx 672 L = 3200 L`.