$n_{C_6H_{12}O_6}=\dfrac{3,5.\dfrac{85}{100}}{180}=\dfrac{119}{7200}(kmol)$ 

`->`$n_{C_6H_{12}O_6(pứ)}=\dfrac{119}{7200}.\dfrac{80}{100}=0,0132(kmol)$ 

$PTHH : C_6H_{12}O_6 \xrightarrow{to,nem-giấm} 2CO_2 +2C_2H_5OH$ 

theo pt 

$n_{C_2H_5OH}=2n_{C_6H_{12}O_6}=0,0264(kmol)$ 

`->m_{C_2H_5OH}=0,0264.46=1,2144(kg)=1214,4(g)`

$V_{C_2H_5OH}=\dfrac{1214,4}{4,8}=253(ml)$ 

ta có

`{V_{rượu}}/{V_{dd}} .100=40^0`

`<=> V_{dd}={253.100}/{40}=632,5(ml)`

Đáp án:

$\rm n_{C_6H_{12}O_6}=\dfrac{3,5×85\%}{180}=\dfrac{119}{7200}\ (kmol)$

Phản ứng xảy ra:

$\rm C_6H_{12}O_6\xrightarrow{men,\ t^°}2C_2H_5OH+2CO_2$

Theo phản ứng:

$\rm n_{C_2H_5OH}=80\%×2n_{C_6H_{12}O_6}=\dfrac{119}{4500}\ (kmol)$

$\rm m_{C_2H_5OH}=\dfrac{119}{4500}×46×1000=1216,44=\ (g)$

$\rm V_{C_2H_5OH}=\dfrac{1216,44}{4,8}=253,425\ (mL)$

$\rm V_{ruou\ (40^°)}=\dfrac{253,425}{40\%}=633,5625\ (mL)$