Đốt cháy hoàn toàn 2,38 gam hỗn hợp E gồm hai este mạch hở X và Y (MX...

* Hướng dẫn giải

\(\begin{array}{l} + \,\,E \to \left\{ \begin{array}{l} COO:\,\,x\,\,mol\\ C{H_2}:\,\,y\,\,mol\\ {H_2}:\,\,z\,\,mol \end{array} \right\} \Rightarrow \left\{ \begin{array}{l} {m_E} = 44x + 14y + 2z = 2,38\\ {n_{C{O_2}}} = x + y = 0,08\\ BTE:\,\,6y + 2z = 0,075.4 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = 0,04\\ y = 0,04\\ z = 0,03 \end{array} \right.\\ + \,\,2,7\,\,gam\,\,muoi\,\,Z \to \left\{ \begin{array}{l} COONa:\,\,0,04\,\,mol\\ C:\,\,a\,\,mol\\ H:\,\,b\,\,mol \end{array} \right\} \to \left\{ \begin{array}{l} N{a_2}C{O_3}:\,\,0,02\,\,mol\\ C{O_2}:\,\,0,02\,\,mol\\ {H_2}O \end{array} \right\} \Rightarrow \left\{ \begin{array}{l} a = 0\\ b = 0,02 \end{array} \right.\\ \Rightarrow Z\,\,gom\,\,\left\{ \begin{array}{l} HCOONa:\,\,0,02\,\,mol\\ {(COONa)_2}:\,\,0,01\,\,mol \end{array} \right\} \Rightarrow \left\{ \begin{array}{l} ancol(0,04\,\,mol)\\ {m_{ancol}} = 2,38 + 0,04.40 - 2,7 = 1,28 \end{array} \right. \Rightarrow {M_{ancol}} = \frac{{1,28}}{{0,04}} = 32\,\,(C{H_3}OH)\\ \Rightarrow E\,\,gom\left\{ \begin{array}{l} X:HCOOC{H_3}:\,\,0,02\,\,mol\\ Y\,:\,\,{(COOC{H_3})_2}:\,\,0,01\,\,mol \end{array} \right\} \Rightarrow \% X = \frac{{0,02.60}}{{2.38}} = \,50,42\% \, \end{array}\)