Đáp án B.
Hỗn hợp $X\left\{ \begin{array}{l} {n_{HCOOC{H_3}}} = x\\ {n_{C{H_3}COO{C_6}{H_5}}} = y \end{array} \right. \to \left\{ \begin{array}{l} \frac{x}{y} = \frac{1}{2}\\ 60x + 136y = 16,6 \end{array} \right. \to \left\{ \begin{array}{l} x = 0,05\\ y = 0,1 \end{array} \right.$
$\begin{align}
& HCOOC{{H}_{3}}+NaOH\to HCOONa+C{{H}_{3}}OH \\
& \,\,\,\,\,\,0,05\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,05\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,05\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \\
\end{align}$
$\begin{align}
& C{{H}_{3}}COO{{C}_{6}}{{H}_{5}}+2NaOH\to C{{H}_{3}}COONa+{{C}_{6}}{{H}_{5}}ONa+{{H}_{2}}O \\
& \,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \\
\end{align}$
Chất rắn gồm:
$\begin{align} & \left\{ \begin{align} & HCOONa\,\,\left( 0,05\,mol \right) \\ & C{{H}_{3}}COONa\,\,\left( 0,1\,mol \right) \\ & {{C}_{6}}{{H}_{5}}ONa\,\,\left( 0,1\,mol \right) \\ & NaOH\,\,d\,\,\left( 0,3-0,05-0,2=0,05\,\,mol \right) \\ \end{align} \right. \\ & \to m=0,05.68+0,1.82+0,1.116+0,05.40=25,2\,\,gam \\ \end{align}$
