`a)` `PTHH:CH_3COOH+NaOH -> CH_3COONa+H_2O` 

`b)` `n_{NaOH}=0,075.2=0,15(mol)`

`n_{CH_3COOH}=n_{NaOH}=0,15(mol)`

`m_{CH_3COOH}=0,15.60=9(g)`

`m_{C_2H_5OH}=18,2-9=9,2(g)`

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

$C_2H_5OH+CH_3COOH \buildrel{{t^o,H_2SO_4đ}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O$

`{0,15}/{1}<{0,2}/{1} -> C_2H_5OH dư`

`n_{CH_3COOC_2H_5}=n_{CH_3COOH}=0,15(mol)`

`m_{CH_3COOC_2H_5}tt=0,15.88.90\%=11,88(g)`

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

 $\rm b.n_{NaOH}=0,075.2=0,15(mol)$

$\rm n_{CH_3COOH}=n_{NaOH}=0,15(mol)$

$\rm m_{CH_3COOH}=0,15.60=9(g)$

$\rm m_{C_2H_5OH}=18,2-9=9,2(g)$

$\rm c.n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2(mol)$

$\rm C_2H_5OH+CH_3COOH \buildrel{{t^o,H_2SO_4đ}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O$

$\rm \dfrac{0,15}{1}<\dfrac{0,2}{1} \xrightarrow{} C_2H_5OH dư$

$\rm n_{CH_3COOC_2H_5}=n_{CH_3COOH}=0,15(mol)$

$\rm m_{CH_3COOC_2H_5}tt=0,15.88.90\%=11,88(g)$