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.

When 50 grams of sulphur was burnt in air, 4% of the impure residue is

1.	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.
Answer» Solution :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`