`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

Options

A. `(1728)/(112)`
B. `(1528)/(73)`
C. `(1920)/(120)`
D. `(112)/(1728)`

Answer

A

Explanation

`{:(,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`