Đáp án:

$\rm a)$

Phản ứng xảy ra:

$\rm C_2H_5OH+CH_3COOH\overset{H_2SO_4\ (dac),\ t^°}\leftrightarrows CH_3COOC_2H_5+H_2O$

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

Theo phản ứng:

$\rm n_{CH_3COOH}=n_{C_2H_5OH}=0,1\ (mol)$

$\rm V_{CH_3COOH}=\dfrac{0,1}{0,5}=0,2\ (M)$

$\rm c)$

Theo phản ứng:

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

$\rm H=\dfrac{7,04}{0,1×88}×100\%=80\%$

` a)`

$\rm PTHH: C_2H_5OH+CH_3COOH \overset{H_2SO_4\ (đ),\ to}\leftrightarrows CH_3COOC_2H_5+H_2O$

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

` b)`   

Theo pthh 

$\rm n_{CH_3COOH}=n_{C_2H_5OH}=0,1\ (mol)$

`->`$\rm V_{CH_3COOH}=\dfrac{0,1}{0,5}=0,2\ (M)$

` c)`

Theo pthh 

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

`->`$\rm H=\dfrac{7,04}{0,1×88}×$`100%=80%`