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A mixture of ethane (C(2)H(6)) and ethen...

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

Text Solution

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Let ethane `=x` Litres
Let ethen `=(40-x)` Litres
`C_(2)H^(6)+C_(2)H_(4)+(13)/(2)O_(2)rarr 4CO_(2)+5H_(2)O`
`(i)` `underset ( (x))C_(2)H_(6)+underset(((7)/(2)x))((7)/(2)O_(2))rarr 2CO_(2)+3H_(2)O`
`(ii).` `underset((40-x)3(40-x))(C_(2)H_(4)+3O_(2))rarr2CO_(2)+2H_(2)O`
Total `O_(2)=(7)/(20x+3(40-x)` at `STP`. ltbr. `[` Total `O_(2)` at `1.00atm` and `400K]`
`PV=nRT`
`n=(PV)/(RT)=(1)/(RT)[(7)/(2)x+3(40-x)]`
Mass of `O_(2)=(1)/(RT)[(7)/(2)x+3(40-X)]xx32` ltbr. `=130 gm (` given `)`
`([(7)/(2)x+3(40-x)]xx32)/(0.082x400)=130`
`x=26.5 `Litre
Percentage of `C_(2)H_(6)=(26.5)/(40)xx100=66.25%`
Percentage of `C_(2)H_(4)=100-66.25=33.75%`
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