Giúp em bài 5 vs đang cần gấp

`n_{CH_3COOH}= {30}/{60}=0,5` (mol)

`n_{ C_2H_5OH}={92}/{46}=2` (mol)

$PTHH: CH_3COOH + C_2H_5OH \buildrel{{\rm \ H_2SO_4 \ đặc , t^o}}\over\longrightarrow CH_3COOC_2H_5+H_2O$ 

Lập tỉ lệ: `{0,5}/1 < 2/1` → `C_2H_5OH` dư

Theo pt : `n_{CH_3COOC_2H_5}` $ (lt)$ `= n_{CH_3COOH}= 0,5` (mol)

Vì `H= 90%`

nên `n_{CH_3COOC_2H_5}` $(tt)$ `= 0,5.90% =0,45` (mol)

`→ m_{CH_3COOC_2H_5}` $(tt)$ ` =0,45.88=39,6`(g)

`n_(CH_3COOH` `=(30)/(60)=0,5(mol)`

`n_(C_2H_5OH` `=(92)/(46)=2(mol)`

 `CH_3COOH+C_2H_5OH` $\buildrel{{H_2SO_{4(\text{đặc})}, t^o}}\over\rightleftharpoons$ `CH_3COOC_2H_5+H_2O`

Do `n_{CH_3COOH}<n_{C_2H_5OH}->` Tỉnh theo `CH_3COOH`

Theo pt: `n_{CH_3COOC_2H_5}=n_{CH_3COOH(\text{pư})}=0,5.60%=0,3(mol)`

`m_(CH_3COOC_2H_5` `=0,3.88=26,4(g)`