Question

what volume of air at NTP containing 21% O2 by volume is required to completely burn 1000 gram of sulphur containig 4% incombustble?

Answer 1

 SO2+ half O2 ------> SO3 22.4 l requires 11.2 l oxygen gas 294 cc will require 147 cc of O2 gas
 let air volume= x cm cube
 0.21*x=147 x=700 cm cube
If the answer is not what you want then consider O2 as O atoms and solve

Answer 2

Answer:

Explanation:

+ 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.