$\rm PTHH :CH_3COOH + NaOH \xrightarrow{} CH_3COONa +H_2O$ 

`n_{NaOH }=0,1.2=0,2(mol)`

theo pthh 

$n_{CH_3COOH}=n_{NaOH}=0,2(mol)$ 

`->`$m_{CH_3COOH}=0,2.60 =12(g)$ 

`=>`$\% m_{CH_3COOH}=\dfrac{12}{16,6}.100\%=72,3\%$ 

`=>`$\% m_{C_2H_5OH}=100\%-72,3\%=27,7\%$ 

$n_{C_2H_5OH}=\dfrac{16,6-12}{46}=0,1(mol)$ 

`b)`

$\rm PTHH: CH_3COOH +C_2H_5OH\buildrel{{H_2SO_4(đ),to}}\over\rightleftharpoons CH_3COOC_2H_5 +H_2O$

tỉ lệ $\dfrac{0,2}{1}>\dfrac{0,1}{1}$ suy ra `C_2H_5OH` hết 

theo pthh $n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,1(mol)$ 

`->`$m_{CH_3COOC_2H_5}=0,1.88=8,8(g)$ 

mà $H=80\%$ 

`->`$m_{CH_3COOC_2H_5(tt)}=8,8.80\%=7,04(g)$ 

Đáp án:

$\rm a)\\n_{NaOH}=0,1×2=0,2\ (mol)$

Phản ứng xảy ra:

$\rm CH_3COOH+NaOH\to CH_3COONa+H_2O$

Theo phương trình:

$\rm n_{CH_3COOH}=n_{NaOH}=0,2\ (mol)$

Phần trăm khối lượng:

$\rm\%m_{CH_3COOH}=\dfrac{0,2×60}{16,6}×100\%=72,3\%\\\%m_{C_2H_5OH}=(100-72,3)\%=27,7\%$

$\rm b)\\n_{C_2H_5OH}=\dfrac{16,6×27,7\%}{46}=0,1\ (mol)$

Phản ứng xảy ra:

$\rm CH_3COOH+C_2H_5OH\overset{H_2SO_4\ (dac),\ t^°}\leftrightarrows CH_3COOC_2H_5+H_2O$

Nhận thấy: $\rm n_{CH_3COOC_2H_5}>n_{C_2H_5OH}\to$ Tính theo $\rm C_2H_5OH$.

$\rm n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,1\ (mol)$

Khối lượng ester thực tế thu được:

$\rm m_{CH_3COOC_2H_5}=0,1×88×80\%=7,04\ (g)$