Question

What volume of 21% oxygen by volume is required of air at NTP to completely burn 1000 gram of sulphur containing 4% incombustible 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.