Giải thích các bước giải:

\(\begin{array}{l}
Zn + 2C{H_3}COOH \to {(C{H_3}COO)_2}Zn + {H_2}\\
a)\\
{n_{Zn}} = 0,2mol\\
 \to {n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
 \to {V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
{n_{C{H_3}{\rm{COO}}H}} = 2{n_{Zn}} = 0,4mol\\
 \to {m_{C{H_3}{\rm{COO}}H}} = 0,4 \times 60 = 24g\\
 \to C{\% _{C{H_3}{\rm{COO}}H}} = \dfrac{{24}}{{200}} \times 100\%  = 12\% \\
c)\\
{C_2}{H_5}OH + C{H_3}COOH \to C{H_3}COO{C_2}{H_5} + {H_2}O
\end{array}\)

\(\begin{array}{l}
{n_{{C_2}{H_5}OH}} = {n_{C{H_3}COOH}} = 0,4mol\\
{m_{{C_2}{H_5}OH}} = 18,4g\\
 \to {V_{{C_2}{H_5}OH}} = \dfrac{{18,4}}{{0,8}} = 23c{m^3} = 2300ml\\
 \to Độ rượu= \dfrac{{50}}{{2300}} \times 100 = 2
\end{array}\)