X, Y, Z là 3 este đều hai chức, mạch hở (trong đó X, Y đều no; Z không no chứa một liên kết đôi C=C). Đun nóng 24,16

* Hướng dẫn giải

$\begin{array}{l}+\left\{\begin{array}{l}{n}_{andehit}=n_{ancolT}=n_{COO}=n_{NaOHpư}=0,36\\ n_{Ag}=1,32\end{array}\right.\Rightarrow 2<\frac{n_{Ag}}{n_{andehit}}<4\Rightarrow \left\{\begin{array}{l}HCHO:0,3mol\\ C{H}_{3}CHO:0,06mol\end{array}\right.\Rightarrow \left\{\begin{array}{l}C{H}_{3}OH:0,3mol\\ C_2{H}_{5}OH:0,06mol\end{array}\right.\\ +\left\{\begin{array}{l}{\pi }_{(X,Y)}=2\stackrel{cắt}{\to }\underset{\pi =2}{\underbrace{2COO}}+\underset{\pi =0}{\underbrace{C_n{H}_{2n+2}}}\stackrel{cắt}{\to }COO+C{H}_2+H_2\Rightarrow n_Y=n_{H_2}\\ {\pi }_Y=3\stackrel{cắt}{\to }\underset{\pi =2}{\underbrace{2COO}}+\underset{\pi =1}{\underbrace{C_m{H}_{2m}}}\stackrel{cắt}{\to }COO+C{H}_2\end{array}\right.\end{array}$

$\begin{array}{l}+E\stackrel{cắt}{\to }\left\{\begin{array}{l}COO:0,36mol\\ C{H}_2:xmol\\ H_2:ymol\end{array}\right.\stackrel{0,92mol{O}_2}{\to }\left\{\begin{array}{l}C{O}_2\\ H_{2}O\end{array}\right.\Rightarrow \left\{\begin{array}{l}0,36.44+14x+2y=24,16\\ BTE:6x+2y=0,92.4\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,58\\ y=0,1\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}{n}_{(X,Y)}=0,1\\ n_Z=0,36-0,1.2=0,08\end{array}\right.\Rightarrow gốcaxit\left\{\begin{array}{l}OOC\mathrm{...}COO:0,1\\ OOC\mathrm{...}CH=CH\mathrm{...}COO:0,08\end{array}\right.\\ \Rightarrow \Delta C{H}_2=0,94-0,3-0,06.2-0,1.2-0,08.4=0\Rightarrow ZlaøC{H}_{3}OOCCH=CHCOOC{H}_3\Rightarrow \boxed{\%Z=47,68\%}\end{array}$

Chọn B