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For the reaction C(2)H(6)(g)hArrC(2)H(4)...

For the reaction `C_(2)H_(6)(g)hArrC_(2)H_(4)(g)+H_(2)(g)`
`K_(p)` is `5xx10^(-2)` atm. Calculate the mole per cent of `C_(2)H_(6)(g)` at equilibruium if pure `C_(2)H_(6)` at `1` atm is passed over a suitable catalyt at `900 K` :

A

`20`

B

`33.33`

C

`66.66`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
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