Câu hỏi 41

\({n_{e\,ste}} = 0,6mol \Rightarrow R - C{\text{OONa:0,6mol}}\)

\({{\text{n}}_{N{a_2}C{O_3}}} = 0,5mol;\, \Rightarrow \)Este của phenol: 0,4mol

${n_{C{O_2}}} = 2,5mol$

\(\begin{gathered}
BTNT\,\,C:0,6(n + 1) + 0,4(m + 6) = 3 \hfill \\
\Leftrightarrow 0,6n + 0,4m = 0 \hfill \\
\Rightarrow \left\{ \begin{gathered}
n = 0 \hfill \\
m = 0 \hfill \\
\end{gathered} \right. \Rightarrow \left\{ \begin{gathered}
HCOONa:0,6 \hfill \\
{C_6}{H_5}ONa:0,4 \hfill \\
\end{gathered} \right.mol \hfill \\
\Rightarrow {m_{{H_2}O}} = 23,4g \hfill \\
\end{gathered} \)