When 50 gm of a sample of sulphur was burnt in air 4% of the sample was left over. Calculate the volume of air required at STP containing 21% oxygen by volume.

Explanation

Weight of sample of sulphur taken =50g
Percentage of impurity =4%
Weight of impurity in sample `=(4 xx 50)/(100)=2gr`
Weight of sulphur in sample =50-2=49grams
Combustion of sulphur gives sulphurdioxide
`S+O_(2) to SO_(2)`
1 mole of S=1 mole of `O_(2)`
32grams of S=22.4 L of `O_(2)` at STP
48 grams of S=?
Volume of oxygen required at STP `=(48)/(32) xx 22.4=33.6 L`
21 L of `O_(2)` is present in 100L of air
33.6 L of `O_(2)` present in ?
Volume of air required at STP `=(33.6)/(21) xx 100=160L`