Question

what volume of air at N.T.P containing 21% of oxygen by volume is required to completely burn 1000g of sulphur containing 4% in combustible matter?

Answer 1

S + O2 → SO2
1mol S reacts with 1 mol O2

You have 1000g S with 4% impurity:
Mass of pure S = 96/100*1000 = 960g S

Molar mass S = 32g/mol
960g = 960/32 = 30 mol S

From the balanced equation you require 30 mol O2
At STP , 1 mol O2 = 22.4L
30 mol O2 = 30*22.4 = 672 L of O2

Volume of air required = 100/21*672 = 3200 L air required.

Answer 2

Answer:

S + O2 → SO2  

1mol S reacts with 1 mol O2  

You have 1000g S with 4% impurity:  

Mass of pure S = 96/100*1000 = 960g S  

Molar mass S = 32g/mol  

960g = 960/32 = 30 mol S  

From the balanced equation you require 30 mol O2  

At STP ,

1 mol O2 = 22.4L  

30 mol O2 = 30*22.4 = 672 L of O2  

Volume of air required = 100/21*672 = 3200 L air required.