Đốt cháy hoàn toàn 6,46 gam hỗn hợp E gồm ba este no, mạch hở X, Y, Z (đều tạo bởi axit cacboxylic và ancol

* Hướng dẫn giải

$+\left\{\begin{array}{l}BTKL:\underset{6,46}{\underbrace{m_E}}+32\underset{0,235}{\underbrace{n_{O_2}}}=44\underset{0,24}{\underbrace{n_{C{O}_2}}}+18\underset{?}{\underbrace{n_{H_{2}O}}}\\ BTO:n_{O/E}+2\underset{0,235}{\underbrace{n_{O_2}}}=2\underset{0,24}{\underbrace{n_{C{O}_2}}}+\underset{?}{\underbrace{n_{H_{2}O}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{H_{2}O}=0,19\\ n_{O/E}=0,2\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{-COO-}=0,1\\ n_{-COONa}=0,1\\ n_{NaOHdư}=20\%.0,1=0,02\end{array}\right.$

$\begin{array}{l}+T\stackrel{quy\text{\hspace{0.17em}\hspace{0.17em}đổ}i}{\to }\left\{\begin{array}{l}COONa:0,1\\ NaOH:0,02\\ C\\ H\end{array}\right\}\stackrel{O_2,t^o}{\to }\left\{\begin{array}{l}BTNa:N{a}_{2}C{O}_3:0,06mol\\ H_{2}O:0,01mol\\ C{O}_2:?\end{array}\right\}\Rightarrow n_{H/H_{2}O}=n_{H/NaOH}\\ \Rightarrow GốcaxitkhôngcóH\Rightarrow Muốilà {\left(COONa\right)}_2:0,05mol.\end{array}$

+ Mặt khác: X, Y, Z no, mạch hở => 2 ancol  phải là đơn chức

$\begin{array}{l}\Rightarrow \left\{\begin{array}{l}{n}_{2ancol}=n_{-COO-}=0,1\\ {\overline{C}}_{ancol}=\frac{6,46+0,1.40-0,05.134}{0,1}=36,7\end{array}\right.\\ \Rightarrow 2ancollà\left\{\begin{array}{l}C{H}_{3}OH\\ C_2{H}_{5}OH\end{array}\right\}\Rightarrow \left\{\begin{array}{l}Xlà{\left(COOC{H}_3\right)}_2(M=118)\\ YlàC{H}_3{\left(COO\right)}_2{C}_2{H}_5(M=132\\ Zlà{\left(COO{C}_2{H}_5\right)}_{2 }(M=146)\end{array}\right\}\Rightarrow \boxed{M_Z=146}\end{array}$

Chọn C