20.2mL of `CH_(6)CO OH` reacts with 201. mL of `C_(2) H_(5) OH` to form `CH_(3) CO OC_(2) H_(5)` (d = 0 . 902 g/mL) by the following reaction
`CH_(3) CO OH+ C_(2) H_(5) OH to CH_(3) CO OC_(2) H_(5) + H_(2) O`
(a) Which compound is the limiting reagent ?
If 27.5 mL of pure ethyl acetate is produced, what is the per cent yield ? Densities of `CH_(3) CO OH and C_(2) H_(5) OH` are 1 . 0 5 g / mL and 0 . 789 g / mL respectively .

Explanation

(a) `CH_(3) C O O H +C_(2) H_(5) OH to CH_(3) CO OH_(2)H_(5) + H_(2) O`
Mole of `CH_(3) CO OH = (2 0.2 xx 1.05)/( 60) =0.3535`
Mole of `C_(2) H_(5) OH = (20 . 1 xx 0.789)/( 46)= 0.3447`
As `CH_(3) CO OH` reacts with `C_(2) H_(5) OH` in a 1 : 1 mole ratio and mole of `C_(2) H_(5)OH` is less than that of `CH_(3) CO OH, C_(2)H_(5)OH` is the limiting reagent.
(b) As ` C_(2)H_(5)OH` is the limiting reagent, mole of `CH_(3) CO OC_(2)H_(15)` to be produced theoretically
= 0.3447 mole
But experimental yield of `CH_(3) CO OC_(2)H_(5) =(27.5 xx 0.902)/(84)`
= 0.2953 mole
`:.` per cent yield of ethyl acetate = `(0.2953)/( 0.3447) xx 100`
` = 85 . 66 %`