1.

A mixture of ethane (C_(2)H_(6)) and ethene (C_(2)H_(4)) occupies 40 litres at 1.00 atm and at 400K. The mixture reacts completely with 130 g of O_(2) to produce CO_(2) and H_(2)O. Assuming ideal gas behaviour, calculate the mole fractions of C_(2)H_(4) and C_(2)H_(6) in the mixture.

Answer»


Solution :Apply PV = nRT and calculate n of the mixture. We get n = 1.22 moles.
Suppose `C_(2)H_(6)="x mole. Then "C_(2)H_(4)=(1.22-x)" mole"`
`2C_(2)H_(6)+7O_(2) rarr 4CO_(2)+6H_(2)O`
`C_(2)H_(4)+3O_(2) rarr 2CO_(2)+2H_(2)O`
`O_(2)" used for x mole of "C_(2)H_(6)=(7)/(2)XX x=3.5x" mole"`
`O_(2)` used for `(1.22-x)` mole of `C_(2)H_(4)=3(1.22-x)` mole
Total `O_(2)` used `=3.5x+3(1.22-x)=(130)/(32)"(GIVEN)"`
Calculate x. We get x = 0.805
`"Mole fraction of "C_(2)H_(6)=(0.805)/(1.22)=0.66`


Discussion

No Comment Found

Related InterviewSolutions