A mixture of ethane and ethene occupies 41 L at atm and 500 K. The mixture reacts compeletly with 10/3 mole of oxygen to produce `CO_(2)` and water. The mole fraction of ethane and ethene in the mixture are (R=0.0821L atm `K^(-1)mol^(-1)` respectively

Options

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

Answer

C

Explanation

`C_(2)H_(6) + 7/2 O_(2) to 2CO_(2) + 3H_(2)O`
`C_(2)H_(4) + 3O_(2) to 2CO_(2) + 2H_(2)O`
Moles of mixture, `n=(pV)/(RT)`
`=(1 "atm" xx 41 L)/((0.082 "L atom" K^(-1) "mol"^(-1)) xx 500 K)`
Let moles of `C_(2)H_(6) =x`, moles of `C_(2)H_(4) =1-x`
Oxygen required
`7/2x + (1-x)3 = 10/3`
`21 x + 18 - 18 x =20`
3x=2 or x = 0.67 mol
Moles of ethane = 0.67
, Moles of ethane =1- 0.67 = 0.33