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

`n_{CH_3COOH}=\frac{300.24%}{60}=1,2(mol)`

theo pthh 

`n_{C_2H_5OH}=n_{CH_3COOH}=1,2(mol)`

`->m_{C_2H_5OH}=1,2.46=55,2(g)`

`b)`

theo pthh 

`n_{CH_3COOC_2H_5}=n_{CH_3COOH}=1,2(mol)`

`->m_{CH_3COOC_2H_5}=1,2.88=105,6(g)`

`->`$m_{CH_3COOC_2H_5(tt)}=105,6.$`70%``=73,92(g)`

`c)`

$\rm PTHH: C_2H_5OH +O_2 \xrightarrow{enzim} CH_3COOH  +H_2O $

theo pthh 

`n_{CH_3COOH}=n_{C_2H_5OH}=1,2(mol)`

`->m_{CH_3COOH}=1,2.60=72(g)`

`->V_{CH_3COOH}=\frac{72}{1,0025}=71,82(ml)`

`=>`$V_{CH_3COOH(tt)}=$`71,28.85%=53,865(ml)`

`≡>V_{dd-CH_3COOH}=\frac{53,865.100}{5}=1077,3(ml)`

Đáp án:

Giải thích các bước giải:

 `a)`

`mct(_(H_2SO_4)) = (300 * 24%)/(100*60) ~~ 0.012 (mol)`

`C_2H_5OH + CH_3COOH `$\xrightarrow{xt,t^o}$` CH_3COOC_2H_5 + H_2O`

Theo phương trình : `n_(CH_3COOH) = n_(C_2H_5OH) = 0.012 (mol)`

`=> m_(C_2H_5OH) = 0.012 * 46 = 0.552 (gam)`

`b)`

Theo phương trình : `n_(CH_3COOH) = n_(CH_3COOC_2H_5) = 0.012 (mol) = n_(H_2O) ` ( Lý thuyết ) 

`=> n_(CH_3COOC_2H_5) = n_(H_2O) ` thực tế `= 0.0084 (mol)`

`=> m_(CH_3COOC_2H_5) = 0.0084 * 88 = 0.7392 (gam)`

`=> m_(H_2O) = 0.0084 * 18 =  0.1512 (mol)`

`c)`

`C_2H_5OH + O_2 `$\buildrel{{t^o,xt}}\over\rightleftharpoons$`CH_3COOH + H_2O`

Theo phương trình : `n_(C_2H_5OH) = n_(CH_3COOH) = 0.012 (mol) `

`=> m_(CH_3COOH) = 0.012 * 60=0.72(gam)` 

`=>V_(CH_3COOH) = (0.72)/(1.0025) *85%=0.61` 

`=> V_(dd )CH_3COOH = (0.61 * 100)/(5) =12.2 (ml)`