When 50 grams of sulphur was burnt in air, `4%` of the impure residue is left over. Calculate the volume of air required at STP containing `21%` of oxygen by volume.

Explanation

Weight of sample of sulphur taken = 50 g
Percentange of impurity `=4%`
Weight of impurity in sample `=(4xx50)/(100)=2gr`
Weight of sulphur `=50-2=40" grams"`
Combustion of sulphur gives sulphurdioxide
`S+O_(2)rarrSO_(2)`
`"1 mole of S"-="1 mole of"O_(2)`
`"32 grams of S"="22.4 L of "O_(2)"at STP"`
`"48 grams of S"=?`
Volume of oxygen required at STP `=(48)/(32)xx22.4=33.6L`
`"21 L of "O_(2)" is present in 100 L of air"`
`"33.6 of "O_(2)" present in ?"`
`"Volume of air required at STP"=(33.6)/(21)xx100=160L`