Đáp án:
mB=3,96 g
77,27% CH3COOC3H7 ; 22,73% CH3COOH
Giải thích các bước giải:
$\begin{gathered}
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O \hfill \\
C{H_3}COO{C_3}{H_7} + NaOH \to C{H_3}COONa + {C_3}{H_7}OH \hfill \\
2{C_3}{H_7}OH + 2Na \to 2{C_3}{H_7}ONa + {H_2} \hfill \\
\end{gathered} $
${n_{{H_2}}} = \dfrac{{0,336}}{{22,4}} = 0,015mol;{n_{NaOH}} = 0,09.0,5 = 0,045mol$
Theo PTHH: ${n_{{C_3}{H_7}OH}} = 2{n_{{H_2}}} = 0,015.2 = 0,03mol$
$\begin{gathered}
\Rightarrow {n_{C{H_3}COO{C_3}{H_7}}} = {n_{{C_3}{H_7}OH}} = 0,03mol \hfill \\
\Rightarrow {n_{C{H_3}COOH}} = 0,045 - 0,03 = 0,015mol \hfill \\
\Rightarrow {m_{C{H_3}COO{C_3}{H_7}}} = 0,03.102 = 3,06g \hfill \\
{m_{C{H_3}COOH}} = 0,015.60 = 0,9g \hfill \\
\Rightarrow {m_B} = 3,06 + 0,9 = 3,96 \hfill \\
\Rightarrow \% {m_{C{H_3}COO{C_3}{H_7}}} = \dfrac{{3,06}}{{3,96}}.100\% = 77,27\% \hfill \\
\% {m_{C{H_3}COOH}} = 100 - 77,27 = 22,73\% \hfill \\
\end{gathered} $