$\rm n_{NaOH}=0,2.0,5=0,1(mol)$

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

$\rm a.n_{CH_3COOH}=n_{NaOH}=0,1mol$

$\rm m_{CH_3COOH}=0,1.60=6(g)$

$\rm \%CH_3COOH=\dfrac{6}{10,6}.100\%=56,6\%$
$\rm \%C_2H_5OH=100-56,6=43,4\%$

$\rm c.CH_3COOH+C_2H_5OH \buildrel{{H_2SO_4đ,t^o}}\over\rightleftharpoons CH_3COOC_3H_5+H_2O$

$\rm n_{C_2H_5OH}=\dfrac{10,6-6}{46}=0,1(mol)$

ta có: $\rm n_{C_2H_5OH}=n_{CH_3COOH}=0,1(Mol)$

$\rm n_{CH_3COOC_2H_5}=n_{CH_3COOH}=0,1(mol)$

$\rm H=80\% \xrightarrow{} m_{CH_3COOC_2H_5}=0,1.88.80\%=7,04(g)$

Đáp án:

Giải thích các bước giải:

 `n_(NaOH) = 0.2 * 0.5 = 0.1 (mol)`

`a)`

`CH_3COOH + NaOH -> CH_3COONa + H_2O`
Theo PT : `n_(NaOH) = n_(CH_3COOH) = 0.1 (mol)`

`=> m_(CH_3COOH) = 0.1 * 60 = 6 (gam)`

`=> %CH_3COOH = 6/10.6 * 100% = 56.60% `

`=> %C_2H_5OH) = 100% - 56.60% = 43.40%`

`b)`

`CH_3COOH + C_2H_5OH` $\buildrel{{t^o,xt}}\over\rightleftharpoons$ `CH_3COOC_2H_5 + H_2O` 

Theo PT : `n_(CH_3COOH) = n_(CH_3COOC_2H_5) = 0.1 (mol)`

Số mol thực tế `CH_3COOC_2H_5` thu được là : `0.1 * 80% =0.08 (mol)`

`=> m_(CH_3COOC_2H_5) = 0.08 * 88 = 7.04 (gam)`