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

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

 b)

theo pthh 

$n_{CH_3COOH}=n_{NaOH }=0,2(mol)$ 

`->`$m_{CH_3COOH}=0,2.60=12(g)$ 

`=>`$m_{C_2H_5OH}=25,8-12=13,8(g)$ 

`c)`

$n_{C_2H_5OH}=\dfrac{13,8}{46}=0,3(mol)$ 

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

tỉ lệ $\dfrac{0,2}{1}<\dfrac{0,3}{1}$ `-> CH_3COOH` hết 

theo pthh 

$n_{CH_3COOC_2H_5}=n_{CH_3COOH}=0,2(mol)$ 

`->`$m_{CH_3COOC_2H_5}=0,2.88=17,6(g)$ 

ta có $H=60\%$ 

`->`$m_{CH_3COOC_2H_5(tt)}=17,6.60\%=10,56(g)$