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., 1 xx 40 = n x 0. 0 821 xx 400`
Total number of moles `n = (1 xx 0)/( 0. 0 821 xx 400) = 1. 2195`
Let x and y be the moles of ethane and ethane, then
`x + y = 1. 1368 " "…(1)`
`C _(2) H _(6) + 7// 2 O _(2) to 2 CO _(2) + 3 H _(2) O , C _(2) H _(4) + 3O _(2) to 2 CO _(2) + 2 H _(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. 0 625" "...(2)`
From equatio (1) and (2), ` x = 0.808, y = 0.41 5`
Mole fraction of ` C_(2) H_(6) = 0.808 //1.2195 = 0.66`
Mole fraction of `C _(2) H _(4) = (1-0.66) = 0.34`