A mixture of hydrocarbon C_(2)H_(2).C_(2)H_(4) & CH_(4) in mole ration of 2:1:2 is burnt completely in the pressence of air containing 80%N_(2) % 20% O_(2) by volume. The mass of air required for the complete combustion of the one gm of mixture is

A mixture of hydrocarbon C_(2)H_(2).C_(2)H_(4) & CH_(4) in mole ration

1.	A mixture of hydrocarbon C_(2)H_(2).C_(2)H_(4) & CH_(4) in mole ration of 2:1:2 is burnt completely in the pressence of air containing 80%N_(2) % 20% O_(2) by volume. The mass of air required for the complete combustion of the one gm of mixture is
Answer» `(1728)/(112)` `(1528)/(73)` `(1920)/(120)` `(112)/(1728)` Solution :`{:(,C_(2)H_(2),:,C_(2)H_(4),:,CH_(4)),(,2,:,1,:,2),(,2x,:,x,:,2x):}` `52x+28x+32x=1` `112x=1` `x=(1)/(112)` `{:(C_(2)H_(2)+(5)/(2)O_(2)rarr2CO_(2)+H_(2)O(l)),((2)/(112)""(5)/(112)),(CH_(4)+3O_(2)rarr2CO_(2)+2H_(2)O(l)),((1)/(112)""(4)/(112)),(CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O(l)),((2)/(112)""(4)/(112)):}` Total moles of`O_(2) = (5)/(112)+(3)/(112)+(4)/(112) = (12)/(112)` moles of `O_(2)= (12)/(112)xx32 = (384)/(112)gm` moles of `N_(2) =(12)/(112)xx4 = (48)/(112)` moles of `N_(2) = (48)/(112)xx28=(1344)/(112)gm` mass of air `=(48+1344)/(112) = (1728)/(112) gm`