Cho 7,34 gam hỗn hợp E gồm hai este mạch hở X và Y...

B

- Hướng dẫn giải

\(\begin{gathered} {n_{COO}} = {n_{NaOH}} = {n_{ancol}} = 2{n_{{H_2}}} = 2.0,05 = 0,1\,mol \hfill \\ BTKL:\,\,{m_{ancol}} = {m_E} + {m_{NaOH}} - {m_T} = 7,34 + 0,1.40 - 6,74 = 4,6g \hfill \\ \Rightarrow {M_{anol}} = 46({C_2}{H_5}OH) \hfill \\ BT\,\,Na:\,\,{n_{N{a_2}C{O_3}}} = \frac{{{n_{NaOH}}}}{2} = 0,05mol \hfill \\ {n_{N{a_2}C{O_3}}} + {n_{C{O_2}}} = 0,05 + 0,05 = 0,1 = {n_{COO}} \Rightarrow T\left\{ \begin{gathered} {(COONa)_2} \hfill \\ HCOONa \hfill \\ \end{gathered} \right. \hfill \\ \Rightarrow E\left\{ \begin{gathered} X:\,\,HCOO{C_2}{H_5} \hfill \\ Y:\,\,{(COO{C_2}{H_5})_2} \hfill \\ \end{gathered} \right. \hfill \\ \left\{ \begin{gathered} {n_{HCOONa}} + 2{n_{{{(COONa)}_2}}} = 0,1 \hfill \\ 68{n_{COONa}} + 134{n_{{{(COONa)}_2}}} = 6,74 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} {n_{HCOONa}} = 0,04 \hfill \\ {n_{{{(COONa)}_2}}} = 0,03 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} {n_X} = 0,04 \hfill \\ {n_Y} = 0,03 \hfill \\ \end{gathered} \right. \hfill \\ \Rightarrow \% X = \frac{{0,04.74}}{{7,34}}.100 = 40,33\% \hfill \\ \end{gathered} \)