Đáp án:

 3,52g và 9,6 g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
n{H_2} = \dfrac{V}{{22,4}} = \dfrac{{2,688}}{{22,4}} = 0,12\,mol\\
 \Rightarrow nhh = 0,12 \times 2 = 0,24\,mol\\
hh:C{H_3}COOH(a\,mol),{C_2}{H_5}OH(b\,mol)\\
60a + 46b = 13,84\\
a + b = 0,24\\
 \Rightarrow a = 0,2;b = 0,04\\
\% m{C_2}{H_5}OH = \dfrac{{0,04 \times 46}}{{13,84}} \times 100\%  = 13,29\% \\
\% mC{H_3}COOH = 100 - 13,29 = 86,71\% \\
b)\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
nC{H_3}COO{C_2}{H_5} = nC{H_3}COO{C_2}{H_5} = 0,04\,mol\\
nC{H_3}COOH = 0,2 - 0,04 = 0,16\,mol\\
mC{H_3}COO{C_2}{H_5} = 0,04 \times 88 = 3,52g\\
mC{H_3}COOH = 0,16 \times 60 = 9,6g
\end{array}\)