Đáp án+ Giải thích các bước giải:

$1)$

$n_{CH_3COOH}$ = `m/M = 3/60 = 0,05(mol)`

$n_{C_2H_5OH}$ = `m/m = {3,68}/46 = 0,08(mol)`

$PT : C_2H_5OH + CH_3COOH$ $\buildrel{{H_2SO_4(đ) , t^o}}\over\rightleftharpoons$ $CH_3COOC_2H_5 + H_2O$

Tỉ lệ : `{0,05}/1` < `{0,08}/1`  `->`  $C_2H_5OH$ dư.

Theo pt : $n_{CH_3COOC_2H_5(LT)}$ = $n_{CH_3COOH}$ = `0,05(mol)`

`->` $m_{CH_3COOC_2H_5(LT)}$ = $n.M = 0,05.88 = 4,4(g)$

`->` $m_{CH_3COOC_2H_5(TT)}$ = $4,4.$`60/100` = $2,64(g)$

$2)$

$n_{CH_3COOC_2H_5(TT)}$ = `m/M = {10,78}/88 = 0,1225(mol)`

$n_{CH_3COOH}$ = `m/M = 12/60 = 0,2(mol)`

$n_{C_2H_5OH}$ = `m/m = {8,05}/46 = 0,175(mol)`

$PT : C_2H_5OH + CH_3COOH$ $\buildrel{{H_2SO_4(đ) , t^o}}\over\rightleftharpoons$ $CH_3COOC_2H_5 + H_2O$

Tỉ lệ : `{0,175}/1` < `{0,2}/1`  `->`  $CH_3COOH$ dư.

Theo pt : $n_{CH_3COOC_2H_5(LT)}$ = $n_{CH_3COOH}$ = `0,175(mol)`

`->` $\text{H%}$ = `{0,1225}/{0,175}` $.100 =$ $\text{70%}$