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

41 L at 1 atm, 500 K = ? L at 1 atm, 273 K (STP)
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
`(1xx41)/(500)=(1xxV)/(273) or V=22.4L=1" mole"`
`C_(2)H_(6)+(7)/(2)O_(2)rarr 2CO_(2)+3H_(2)O`
`C_(2)H_(2)+3O_(2)rarr 2CO_(2)+2H_(2)O`
Suppose `C_(2)H_(6)=x` mole, Then `C_(2)H_(4)=1-x `mole `O_(2)` consumed `=(7x)/(2)+3(1-x)`
`=3+(x)/(2)=(10)/(2)" (given)"`
`"or "x=2((10)/(3)-3)=(2)/(3)=0.67`