Cho 16,6 gam hỗn hợp X gồm metyl fomat và phenyl axetat tác dụng hoàn toàn với 200 ml dung dịch NaOH, tính m?

* Hướng dẫn giải

Hỗn hợp \(X\left\{ \begin{array}{l} {n_{HCOOC{H_3}}} = x\\ {n_{C{H_3}COO{C_6}{H_5}}} = y \end{array} \right. \to \left\{ \begin{array}{l} \frac{x}{y} = \frac{1}{2}\\ 60x + 136y = 16,6 \end{array} \right. \to \left\{ \begin{array}{l} x = 0,05\\ y = 0,1 \end{array} \right.\)

\(\begin{array}{l} HCOOC{H_3} + NaOH \to HCOONa + C{H_3}OH\\ \,\,\,\,\,\,0,05\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,05\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,05\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \end{array}\)

\(\begin{array}{l} C{H_3}COO{C_6}{H_5} + 2NaOH \to C{H_3}COONa + {C_6}{H_5}ONa + {H_2}O\\ \,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \end{array}\)

Chất rắn gồm:

\(\left\{ \begin{array}{l} HCOONa\,\,\left( {0,05\,mol} \right)\\ C{H_3}COONa\,\,\left( {0,1\,mol} \right)\\ {C_6}{H_5}ONa\,\,\left( {0,1\,mol} \right)\\ NaOH\,\,du\,\,\left( {0,3 - 0,05 - 0,2 = 0,05\,\,mol} \right) \end{array} \right. \to m = 0,05.68 + 0,1.82 + 0,1.116 + 0,05.40 = 25,2\,\,gam\)