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

PV = nRT
`n = (1 xx 41)/(0.082 xx 500) = 1 `mole
`underset("1 mole")(C_(2)H_(4)) + underset("3 mole")(3O_(2)) to 2CO_(2) + 2H_(2)O`
`underset("2 mole")(2C_(2)H_(6)) + underset(" 7 moles")(7O_(2)) to 4CO_(2) + 6H_(2)O`
If in the mixture of `C_(2)H_(6)` and `C_(2)H_(4)`, x moles of `C_(2)H_(6)` are present than (1-x) moles of `C_(2)H_(4)` are present .
Total moles of `O_(2)` required = 10/3
`(7)/(2)x + 3(1-x)= (10)/(3)`
= x = 2//3
Moles of ethane = 2/3
Moles of ethene = 1/3
Mole fraction of ethane = (2 / 3)/1 = 2 / 3
Mole fraction of ethence = (1/3)/1 = 1/3.