If `1` mole of `CH_(3)COOH` and `1` mole of `C_(2)H_(5)OH` are taken in `1` litre flask, `50%` of `CH_(3)COOH` is converted into ester as,
`CH_(3)COOH_((l))+C_(2)H_(5)OH_((l))hArrCH_(3)COOC_(2)H_(5)(l)+H_(2)O_((l))`
There is `33%` conversion of `CH_(3)COOH` into ester, if `CH_(3)COOH "and" C_(2)H_(5)OH` have been taken initially in molar ratio `x:1`, find `x`.

Answer

2

Explanation

`CH_(3)COOH_((l))+C_(2)H_(5)OH_((l))hArrCH_(3)COOC_(2)H_(5(l)+H_(2)O_((l))`
`1-0.5 1-0.5 0.5 0.5`
So, `K_(c)=(0.5xx0.5)/(0.5xx0.5)=1`
Now let a moles of `CH_(3)COOH "and" b "moles of" C_(2)H_(5)OH` are taken:
`a-(a)/(3) b-(a)/(3) (a)/(3) (a)/(3)`
So, `K_(c)=((a//3)xx(a//3))/(2a//3xx(b-(a)/(3))) or 2(b-(a)/(3))=(a)/(3)`
or `2b=a` or `(a)/(b)=(2)/(1)`