Đáp án C
$\begin{array}{l} {n_{NaOH(da{\rm{ }}dung)}} = \frac{{150.5\% }}{{40}} = 0,1875mol\\ {n_{NaOH(da{\rm{ }}dung)}} = {n_{NaOH(phan{\rm{ }}ung)}} + {n_{NaOH(du)}}\\ {n_{NaOH(da{\rm{ }}dung)}} = {n_{NaOH(phan{\rm{ }}ung)}} + 25\% .{n_{NaOH(phan{\rm{ }}ung)}}\\ \to {n_{NaOH(phan{\rm{ }}ung)}} = \frac{{{n_{NaOH(da{\rm{ }}dung)}}}}{{1,25}} = 0,15mol\\ 10,4gam\left\{ \begin{array}{l} C{H_3}COOH:x\\ C{H_3}COO{C_2}{H_5}:y \end{array} \right. \end{array}$
Ta có:
$\begin{array}{l} \left\{ \begin{array}{l} x + y = 0,15\\ 60x + 88y = 10,4 \end{array} \right. \to \left\{ \begin{array}{l} x = 0,1\\ y = 0,05 \end{array} \right.\\ \to \% {m_{etylaxetat}} = 42,31\% \end{array}$
