Giải thích các bước giải:

Bài 1:

\(\begin{array}{l}
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
C{H_3}{\rm{COOH + Na}} \to C{H_3}{\rm{COO}}Na + \dfrac{1}{2}{H_2}\\
{n_{{H_2}}} = 0,075mol\\
\left\{ \begin{array}{l}
46a + 60b = 7,6\\
a + \dfrac{1}{2}b = 0,075
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,019\\
b = 0,112
\end{array} \right.\\
 \to {n_{{C_2}{H_5}OH}} = 0,019mol \to {m_{{C_2}{H_5}OH}} = 0,019 \times 46 = 0,874g\\
 \to \% {m_{{C_2}{H_5}OH}} = \dfrac{{0,874}}{{7,6}} \times 100\%  = 11,58\% \\
 \to {n_{C{H_3}{\rm{COO}}H}} = 0,112mol \to {m_{C{H_3}{\rm{COO}}H}} = 0,112 \times 60 = 6,72g\\
 \to {m_{C{H_3}{\rm{COO}}H}} = \dfrac{{6,72}}{{7,6}} \times 100\%  = 88,42\% \\
{C_2}{H_5}OH + C{H_3}{\rm{COOH}} \to C{H_3}{\rm{COO}}{C_2}{H_5} + {H_2}O\\
{n_{{C_2}{H_5}OH}} < {n_{C{H_3}{\rm{COO}}H}}\\
 \to {n_{C{H_3}{\rm{COO}}{{\rm{C}}_2}{H_5}}} = {n_{{C_2}{H_5}OH}} = 0,019mol\\
 \to {m_{C{H_3}{\rm{COO}}{{\rm{C}}_2}{H_5}(tt)}} = 0,019 \times 88 = 1,672g\\
 \to H = \dfrac{{1,672}}{{11}} \times 100\%  = 15,2\% 
\end{array}\)

Bài 2:

\(\begin{array}{l}
C{H_3}{\rm{COOH + NaOH}} \to C{H_3}{\rm{COO}}Na + {H_2}O\\
{n_{NaOH}} = 4 \times 0,1 = 0,4mol\\
 \to {n_{C{H_3}{\rm{COO}}H}} = {n_{NaOH}} = 0,4mol\\
 \to {C_{M(}}_{C{H_3}{\rm{COO}}H)} = \dfrac{{0,4}}{{0,05}} = 8M\\
 \to {n_{C{H_3}{\rm{COONa}}}} = {n_{NaOH}} = 0,4mol\\
 \to {C_{M{\rm{dd}}A}} = \dfrac{{0,4}}{{0,1 + 0,05}} = 2,67M
\end{array}\)