When `C_(2)H_(5)OH` and `CH_(3)COOH` are mixed in equivalent proportion, equilibrium is reached when `2//3` of acid and alcohol are used. How much ester will be present when `2g` "mole"cule of acid were to react with `2g` "mole"cule of alcohol.

Explanation

`{:(,C_(2)H_(5)OH,+,CH_(3)COOH,hArr,H_(3)COOC_(2)H_(5),+,H_(2)O),("moles before reaction",1,,1,,0,,0),("moles at equilibrium",1-x,,1-x,,x,,x),(,:' x=2//3,,,,,,),("moles at equilibrium",:. (1-2//3),,(1-2//3),,2//3,,2//3):}`
`K_(c ) =([H_(3)COOC_(2)H_(5)][H_(2)O])/([C_(2)H_(5)OH][CH_(3)COOH])`
`:. K_(c ) =(2//3xx2//3)/(1//3xx1//3)=4`