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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 `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. |
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Answer» For a gaseous mixture of `C_(2)H_(6)` and `C_(2)H_(14)` `PV=nRT` `:. 1xx40=nxx0.082xx400` `:.` Total mole of `(C_(2)H_(6)+C_(2)H_(4))=1.2195` Let mole `C_(2)H_(6)` and `C_(2)H_(4)` be `a` and `b` respectively. `a+b=1.2195`….(1) `C_(2)H_(6)+(7//2)O_(2)rarr2CO_(2)+3H_(2)O` `C_(2)H_(4)+3O_(2)rarr2CO_(2)+2H_(2)O` `:.` Mole of `O_(2)` needed for complete reaction of mixture `=7a//2+3b` `:. (7a)/(2)+3b=(130)/(32)`....(2) By eqs.(1) and (2), `a=0.808, b=0.4115` `:.` Mole fraction of `C_(2)H_(6)-0.808//1.2195=0.066` and Mole fraction of `C_(2)H_(4)=0.34` |
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