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

Case I:
`{:(,C_(2)H_(5)OH,+,CH_(3)COOH,hArr,CH_(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",(1-2//3),,(1-2//3),,2//3,,2//3):}`
`:. K_(c)=((2)/(3)xx(2)/(3))/((1)/(3)xx(1)/(3))=4`
`{:(Case.II:,C_(2)H_(5)OH,+,CH_(3)COOH,hArr,CH_(3)COOC_(2)H_(5),+,H_(2)O),("Moles before reaction",2,,2,,0,,0),("Moles at equilibrium",2-x,,2-x,,x,,x):}`
`:. K_(c)=4=(x^(2))/((2-x)^(2))`
or `(x)/((2-x))=2`
or `x=1.33`