Đáp án:

 a/$\%m_{C_2H_5OH}=56,1\%\\\%m_{CH_3COOH}=43,9\%$

b/ 11,88g

Giải thích các bước giải:

 $Gọi\ n_{C_2H_5OH}=a;n_{CH_3COOH}=b$

$CH_3COOH+Na\to CH_3COONa+\dfrac{1}{2}H_2\\b\hspace{7,5cm}0,5b\\C_2H_5OH+Na\to C_2H_5ONa+\dfrac{1}{2}H_2\\a\hspace{6,5cm}0,5a$

Theo đề bài, ta có hệ phương trình: 

$\begin{cases}46a+60b=20,5\\0,5a+0,5b=\dfrac{4,48}{22,4}\end{cases}⇒a=0,25;b=0,15$

a/ $m_{C_2H_5OH}=0,25.46=11,5g⇒\%m_{C_2H_5OH}=\dfrac{11,5}{20,5}.100\%=56,1\%\\\%m_{CH_3COOH}=100-56,1=43,9\%$

b/ 

$CH_3COOH+C_2H_5OH⇄CH_3COOC_2H_5+H_2O$

+/ Nếu H% = 100%, $n_{este}=n_{CH_3COOH}=0,15\ mol$

+/ Nếu H%=90% ⇒ $n_{este}=0,15.\dfrac{90}{100}=0,135\ mol$

$⇒m_{Este}=0,135.88=11,88g$