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

Amount of Sulphur taken = 50 gm
Wt. of sulphur left = 4% = 2 gm
Wt. of sulphur reacted = 50 - 2 = 48 gm
Sulphur burms in air according to the reaction
`S+O_(2)toSO_(2)`
Moles of Sulphur = `48/32` = 1.5
Moles of Oxygen required = 1.5
Volume of oxygen at STP = `22.4xx1.5=33.6`lit.
Volume of air = `(33.6xx100)/21=160`lit.
`" "(therefore"air is "21%O_(2))`