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`
`1xx41=nxx0.0821xx500rArrn=0.998mol`
The number of moles of ethane ` = x`
So number of moles of ethene `=(0.998-x)`
Reaction of ethane and ethene with `O_(2)`
(i) `2C_(2)H_(6)+70_(2)to4CO_(2)+6H_(2)O`
(ii) `C_(2)H_(4)+3O_(2)to2CO_(2)+2H_(2)O`
x mole of ethane reacts `"with"(7x)/(2)"moles of " O_(2)`
`(0.998-x)` mole of ethene reacts
`=3(0.998-x)"mole of "O_(2)`
`(7x)/(2)+[3(0.998-x)]=(10)/(3)"mole of "O_(2)`
`x=(0.3393)/(0.5)=0.678"mole of ethane"`
`"mole of ethene"=0.998-0.678=0.32`
`x_("enthane")=(eta_("ethane"))/(eta_("ethane")+eta_("ethene"))=(0.678)/(0.678+0.32)=0.67`
`x_("ethene")=1-x_("ethane")=1-0.67=0.33`