20.2mL of CH_(6)CO OHreacts with 201.mL of C_(2) H_(5) OHto formCH_(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 compoundis the limitingreagent ? If 27.5 mL of pureethyl acetate is produced, whatis the per cent yield ? Densities of CH_(3) CO OHand C_(2) H_(5) OH are 1 . 0 5 g / mL and 0 . 789 g / mL respectively .

20.2mL of CH_(6)CO OHreacts with 201.mL of C_(2) H_(5) OHto formCH_(3)

1.	20.2mL of CH_(6)CO OHreacts with 201.mL of C_(2) H_(5) OHto formCH_(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 compoundis the limitingreagent ? If 27.5 mL of pureethyl acetate is produced, whatis the per cent yield ? Densities of CH_(3) CO OHand C_(2) H_(5) OH are 1 . 0 5 g / mL and 0 . 789 g / mL respectively .
Answer» SOLUTION :(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 limitingreagent. (b) As ` C_(2)H_(5)OH` is thelimiting 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 centyieldof ethyl acetate = `(0.2953)/( 0.3447) xx 100` ` = 85 . 66 %`