A mixture of 40 litres of ethane and ethene at 1 atm pressure and 400k reacts completely with `130g" of "O_(2)` to form `CO_(2)` and `H_(2)O`. Assuming ideal behaviour, calculate the mole fraction of ethane and ethene.

Explanation

According to ideal gas equation, PV=nRT, i.e, `1xx40=nxx0.0821xx400`
Total number of moles,`n=(1xx40)/(0.0821xx400)=1.2195`
Let x and y be the moles of ethane and ethene, then
`x+y=1.1368" ""............"(1)`
`C_(2)H_(6)+7"/"2O_(2) rarr 2CO_(2)+3H_(2)O`
`C_(2)H_(4)+3O_(2) rarr 2CO_(2)+2H_(2)O`
Moles of `O_2` needed for complete reaction of the mixture `=7"/"2x+3y`
Number of moles of `O_(2) =130"/"32=4.0625`
`7"/"2x+3y=4.0625" "".........."(2)`
From equation (1) and (2), `x=0.808, y=0.415`
Mole friction of `C_(2)H_(6)= 0.808"/"1.2195=0.66`
Mole fraction of `C_(2)H_(4)=(1.0.66)=0.34`.