Cho 0,06 mol hỗn hợp hai este đơn chức X và Y (M

C

- Hướng dẫn giải

\(\begin{array}{l}
 + \,\,\left\{ \begin{array}{l}
{n_{X,\,\,Y}} = 0,06\\
{n_{KOH}} = 2{n_{{K_2}C{O_3}}} = 0,07
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{n_X} = 0,05\\
{n_Y} = 0,01
\end{array} \right.\\
 + \,\,{\overline C _{X,\,\,Y}} = \frac{{{n_{C{O_2}}} + {n_{{K_2}C{O_3}}}}}{{{n_{X,\,\,Y}}}} = 3 \Rightarrow \left\{ \begin{array}{l}
X:\,\,HCOOC{H_3}\,\,({C_X} = 2)\\
{C_Y} = \frac{{0,18 - 0,05.2}}{{0,01}} = 8 \Rightarrow Y\,:\,HCOO{C_6}{H_4} - C{H_3}
\end{array} \right.\\
 + \,\,\,\left\{ \begin{array}{l}
HCOOK:0,06\\
C{H_3} - \,{C_6}{H_4}OK:0,01
\end{array} \right. \Rightarrow m = 6,50\,g.\,\,
\end{array}\)