`n_{CH_3COOH}=90/60=1,5(mol)`

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

`(mol)`             `1,5->`                 `1,5`                               `1,5`

`b.m_{CH_3COOC_2H_5(l t)}=1,5.88=132(g)`

`H=80%=>m_{CH_3COOC_2H_5} tt=132.80%=105,6(g)`

`c.m_{C_2H_5OH}=1,5.46=69(g)`

`d.` cho `D_{rượu}=0,8g//ml`

`V_{rượu}=0,8.69=55,2(g)`

độ rượu = `{55,2}/{55,2+100}.100~~35,57^o`

Đáp án:

Giải thích các bước giải:

 `a)`

`n_(CH_3COOH) = 90/60 = 1.5 (mol)`

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

`1.5 -> 1.5 -> 1.5 -> 1.5`

`b)`

`m_(CH_3COOC_2H_5) = 1.5 * 88 = 132 (gam)`

`-> m_(CH_3COOC_2H_5) ` thực tế : `132 * 80% = 105.6 (gam)`

`c)`

`m_(C_2H_5OH) = 1.5 * 46 =  69 (gam)`

`d)`

`D_(C_2H_5OH) = 0.8 (gam`/`ml)` 

`-> V_(C_2H_5OH) = 0.8 *69 = 55.2 (gam)`

`-> ` Độ rượu `= (55.2)/(100 + 55.2) *100= 35.57^o`