Em tham khảo nha :

\(\begin{array}{l}
1)\\
{C_6}{H_5}OH + NaOH \to {C_6}{H_5}ONa + {H_2}O\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
{n_{NaOH}} = 0,2 \times 0,5 = 0,1mol\\
hh:{C_6}{H_5}OH(a\,mol);C{H_3}COOH(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,1\\
94a + 60b = 6,85
\end{array} \right.\\
 \Rightarrow a = 0,025;b = 0,075\\
{m_{{C_6}{H_5}OH}} = 0,025 \times 94 = 2,35g\\
{m_{C{H_3}COOH}} = 6,85 - 2,35 = 4,5g\\
2)\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}COOH}} = \dfrac{6}{{60}} = 0,1mol\\
{n_{{C_2}{H_5}OH}} = \dfrac{{9,2}}{{46}} = 0,2mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,2}}{1} \Rightarrow {C_2}{H_5}OH\text{ dư}\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,1mol\\
{m_{C{H_3}COO{C_2}{H_5}}} = 0,1 \times 88 = 8,8g\\
H = \dfrac{{5,5}}{{8,8}} \times 100\%  = 62,5\% 
\end{array}\)

 

Đáp án:

`1)` $\begin{cases} m_{C_6H_5OH}=2,35(g)\\m_{CH_3COOH}=4,5(g) \end{cases}$

`2)` `H=62,5%`

Giải thích các bước giải:

`1)`

`PT : C_6H_5OH + NaOH -> C_6H_5ONa + H_2O`

         `CH_3COOH + NaOH -> CH_3COONa + H_2O`

Đổi `200(ml)=0,2(l)`

`n_{NaOH}=C_MxxV=0,5xx0,2=0,1(mol)`

Gọi $\begin{cases} a:n_{C_6H_5OH}\\b:n_{CH_3COOH} \end{cases}$   `(a,b>0)`

Theo đề , ta có hpt : $\begin{cases} 94a+60b=6,85\\a+b=0,1 \end{cases}$ `=>` $\begin{cases} a=0,025\\b=0,075 \end{cases}$

`=>` $\begin{cases} m_{C_6H_5OH}=n.M=0,025.94=2,35(g)\\m_{CH_3COOH}=n.M=0,075.60=4,5(g) \end{cases}$

`2)`

`n_{CH_3COOH}=m/M=6/60=0,1(mol)`

`n_{C_2H_5OH}=m/M={9,2}/46=0,2(mol)`

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

Tỉ lệ : `{0,1}/1<{0,2}/1=>C_2H_5OH` dư.

Theo pt : `n_{CH_3COOC_2H_5(l t)}=n_{CH_3COOH}=0,1(mol)`

`=>H={m_{CH_3COOC_2H_5(t t)}xx100%}/{m_{CH_3COOC_2H_5(l t)}}={5,5xx100%}/{0,1xx88}=62,5%`

$\text{#Cyrus}$