A mixture of ethane and ethene occupies 41 L at 1 atm and 500 K. The mixture reacts completely with `(10)/(3)` mole of `O_(2)` to produce `CO_(2) and H_(2)O`. The mole fraction of ethane and ethene in the mixture are respectively (R = 0.082 L atm `K^(-1)mol^(-1)`)

Options

A. 0.50, 0.50
B. 0.75, 0.25
C. 0.67, 0.33
D. 0.25, 0.75

Answer

C

Explanation

For a gaseous mixture of `C_(2)H_(6) and C_(2)H_(4)`
pV = nRT
or `n = (pV)/(RT) = (1 xx 41)/(0.082 xx 500)`
or n = 1
`therefore` Total mole of `C_(2)H_(6) + C_(2)H_(4) = 1` mole
Let the mole of `C_(2)H_(6) = x`
then mole of `C_(2)H_(4) = 1 - x`
`C_(2)H_(6) + (7)/(2)O_(2)rarr 2CO_(2) + 3H_(2)O`
`C_(2)H_(4) + 3O_(2)rarr 2CO_(2) + 2H_(2)O`
`therefore` Mole of `O_(2)` needed for complete reaction of mixture
`= (7)/(2) x + 3(1 - x)`
`therefore" "(7)/(2)x + 3(1-x) = (10)/(3)`
or `x = (2)/(3)`
Thus, mole fraction of `C_(2)H_(6) = (2)/(3) = 0.67`
and mole fraction of `C_(2)H_(4) = 1 - (2)/(3) = 0.33`