X, Y là 2 axit cacboxylic đều mạch hở; Z là ancol no; T là este hai chức

* Hướng dẫn giải

Đáp án D

$$\begin{gathered} + E gồm \left\{\begin{array}{l}X là R\text{'}COOH :x mol; Y là R\text{'}\text{'}COOH: y mol\\ Z là R(OH{)}_2: z mol; T là R\text{'}COOROOCR\text{'}\text{'}: t mol\end{array}\right. \\ + Hỗn hợp E\left\{\begin{array}{l}\left\{\begin{array}{l}\underset{(x+t) mol}{\underbrace{R\text{'}COONa}}\\ \underset{(y+t) mol}{\underbrace{R\text{'}\text{'}COONa}}\end{array}\underset{0,5 mol}{\overset{O_2,t^0}{\to }}\right.C{O}_2+\underset{0,2 mol}{\underbrace{N{a}_{2}C{O}_3}}+\underset{0,38 mol}{\underbrace{H_{2}O}}\\ \underset{(z+t) mol}{\underbrace{R{\left(OH\right)}_2}}\stackrel{Na}{\to }\underset{(z+t)=0,26}{\underbrace{H_2\uparrow }}; m_{bình Na tăng}=19,24 gam\end{array}\right. \\ \begin{array}{l}+ \left\{\begin{array}{l}{n}_{R{\left(OH\right)}_2}=n_{H_2}=0,26\\ \underset{19,24}{\underbrace{m_{bình tăng}}}=\underset{?}{\underbrace{m_{R{\left(OH\right)}_2}}}-\underset{0,26.2}{\underbrace{m_{H_2}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{R{\left(OH\right)}_2}=0,26; m_{R{\left(OH\right)}_2}=19,76\\ M_{R{\left(OH\right)}_2}=\frac{19,76}{0,26}=76: C_3{H}_6(OH{)}_2\end{array}\right.\\ + \left\{\begin{array}{l}{n}_{N{a}_{2}C{O}_3}=0,5n_{NaOH}=0,2\\ \underset{0,4.2}{\underbrace{n_{O muối}}}+2\underset{0,5}{\underbrace{n_{O_2}}}=3\underset{0,2}{\underbrace{n_{N{a}_{2}C{O}_3}}}+2\underset{?}{\underbrace{n_{C{O}_2}}}+\underset{0,4}{\underbrace{n_{H_{2}O}}}\end{array}\right.\Rightarrow n_{C{O}_2}=0,4\Rightarrow {\overline{C}}_{muối}=\frac{0,4+0,2}{0,4}=1,5\end{array} \\ \Rightarrow Hai muối là HCOONa và C{H}_{3}COONa \\ \begin{array}{l}+ \left\{\begin{array}{l}{n}_{R\text{'}COONa}=n_{R\text{'}\text{'}COONa}\\ n_{-COO-}=n_{NaOH}\\ n_{R{\left(OH\right)}_2}=n_{H_2}\\ m_E=37,36\end{array}\right.\Rightarrow \left\{\begin{array}{l}x+t=y+t=0,2\\ x+y+2t=0,4\\ z+t=0,26\\ 46x+60y+76z+146t=37,36\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=y\\ 2x+2t=0,4\\ z+t=0,26\\ 106x+76z+146t=37,36\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=y=0,1\\ z=0,16\\ t=0,1\end{array}\right.\\ \Rightarrow \%m_{T \left(HCOO{C}_3{H}_{6}OOCC{H}_3\right)}=\frac{0,1.146}{37,36}.100\%=39,08\%\end{array} \end{gathered}$$