Giải thích các bước giải:
\(\begin{array}{l}
2C{H_3}{\rm{COO}}H + {K_2}C{O_3} \to 2C{H_3}{\rm{COO}}K + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = 0,2mol \to {n_{C{H_3}{\rm{COO}}H}} = 2{n_{C{O_2}}} = 0,4mol\\
{C_2}{H_5}OH + Na \to {C_2}{H_5}ONa + \dfrac{1}{2}{H_2}\\
C{H_3}{\rm{COO}}H + Na \to C{H_3}{\rm{COONa}} + \dfrac{1}{2}{H_2}\\
{n_{{H_2}}} = 0,5mol \to {n_{{C_2}{H_5}OH}} = 2 \times (0,5 - \dfrac{1}{2} \times 0,4) = 0,6mol\\
\to {n_{C{H_3}{\rm{COO}}H}}(A) = 0,8mol \to {m_{C{H_3}{\rm{COO}}H}} = 48g\\
\to {n_{{C_2}{H_5}OH}}(A) = 1,2mol \to {m_{{C_2}{H_5}OH}} = 55,2g\\
C{H_3}{\rm{COO}}H + {C_2}{H_5}OH \to C{H_3}{\rm{COO}}{C_2}{H_5} + {H_2}O\\
{n_{{C_2}{H_5}OH}} > {n_{C{H_3}{\rm{COO}}H}}
\end{array}\)
\({C_2}{H_5}OH\) dư
\(\begin{array}{l}
\to {n_{C{H_3}{\rm{COO}}{C_2}{H_5}}} = {n_{C{H_3}{\rm{COO}}H}} = 0,8mol\\
\to {m_{C{H_3}{\rm{COO}}{C_2}{H_5}}} = 70,4g\\
\to {m_{C{H_3}{\rm{COO}}{C_2}{H_5}}}thực tế= \dfrac{{70,4 \times 60}}{{100}} = 42,24g
\end{array}\)
