Đáp án+ Giải thích các bước giải:

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

`-` `CH_3COOC_2H_5` : Etyl axetat

`n_{CH_3COOH} = m/M = 90/60 = 1,5(mol)`

`n_{C_2H_5OH} = m/M = 46/46 = 1(mol)`

$n_{CH_3COOC_2H_5(TT)}$` = m/M = 55/88 = 0,625(mol)`

Tỉ lệ : `{1,5}/1` > `1/1`  `-> CH_3COOH` dư.

Theo pt : `n_{CH_3COOC_2H_5(LT)} = n_{C_2H_5OH} = 1(mol)`

`-> H% =` $\dfrac{0,625}{1}$.`100(%) = 62,5(%)`

$\text{#Cyrus}$

Đáp án:

`↓`

Giải thích các bước giải:

`PTHH:`         `CH_3COOH+C_2H_5OH` $\buildrel{{H_2SO_{4(đ)}, t^o}}\over\rightleftharpoons$ `CH_3COOC_2H_5+H_2O`

`→CH_3COOC_2H_5:` Etyl axetat

`n_{CH_3COOH}=``\frac{m}{M}=\frac{90}{60}=1,5(mol)`

`n_{C_2H_5OH}=``\frac{m}{M}=\frac{46}{46}=1(mol)`

`PTHH`        `CH_3COOH+C_2H_5OH` $\buildrel{{H_2SO_{4(đ)}, t^o}}\over\rightleftharpoons$ `CH_3COOC_2H_5+H_2O`

Vì `\frac{n_{CH_3COOH}}{1}>\frac{n_{C_2H_5OH}}{1}→CH_3COOH` dư

Theo pt, ta có: `n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=1(mol)`

`→m_{CH_3COOC_2H_5}=n.M=1.88=88(g)`

`⇒H=``\frac{m_{thực tế}.100%}{m_{lí thuyết}}=\frac{55.100%}{88}=62,5%`