What volume of air at n.t.p containing 21% of oxygenby volume is required to completely burn 1000 g of sulphur containing 4% incombustible matter?
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Answered by
45
HEYA......
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.
TYSM.....$GOZMIT
S + O2 → SO2
1mol S reacts with 1 mol O2
You have 1000g S with 4% impurity:
Mass of pure S = 96/100*1000 = 960g S
Molar mass S = 32g/mol
960g = 960/32 = 30 mol S
From the balanced equation you require 30 mol O2
At STP ,
1 mol O2 = 22.4L
30 mol O2 = 30*22.4 = 672 L of O2
Volume of air required = 100/21*672 = 3200 L air required.
TYSM.....$GOZMIT
Answered by
3
Volume of air required is 3200 litres
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