`n_{C_2H_5OH}={4,6}/46=0,1(mol)`

`n_{CH_3COOH}=9/60=0,15(mol)`

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

ta có: `{0,1}/1 <{0,15}/1 =>CH_3COOH` dư

`n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,1(mol)`

`m_{CH_3COOC_2H_5}l t=0,1.88=8,8(g)`

`H=30%=>m_{CH_3COOC_2H_5}=8,8.30%=2,64(g)`

$PTHH  : CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 +H_2O$ 

$n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1(mol)$ 

$n_{CH_3COOH}=\dfrac{9}{60}=0,15(mol)$ 

tỉ lệ `{0,15}/{1}<{0,1}/{1}` suy ra $C_2H_5OH$  hết 

theo pthh 

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

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

`=>`$m_{CH_3COOC_2H_5(tt)}=8,8.\dfrac{30}{100}=2,64(g)$