X là este đơn chức, không no chứa một liên kết đôi C=C; Y là este no, hai chức (X, Y đều mạch hở). Đốt cháy hoàn toàn 10,36

* Hướng dẫn giải

$\begin{array}{l}+n_{NaOHpư}=x\Rightarrow 1,25x=0,15\Rightarrow x=0,12=n_{COO}.\\ +\left\{\begin{array}{l}{\pi }_X=2\stackrel{cắt}{\to }\underset{\pi =1}{\underbrace{COO}}+\underset{\pi =1}{\underbrace{C_n{H}_{2n}}}\stackrel{cắt}{\to }COO+C{H}_2\\ {\pi }_Y=2\stackrel{cắt}{\to }\underset{\pi =2}{\underbrace{2COO}}+\underset{\pi =0}{\underbrace{C_m{H}_{2m+2}}}\stackrel{cắt}{\to }COO+C{H}_2+H_2\Rightarrow n_Y=n_{H_2}\end{array}\right.\\ +E\stackrel{cắt}{\to }\left\{\begin{array}{l}COO:0,12mol\\ C{H}_2:xmol\\ H_2:ymol\end{array}\right.\stackrel{O_2}{\to }\left\{\begin{array}{l}C{O}_2\\ H_{2}O\end{array}\right.\Rightarrow \left\{\begin{array}{l}0,12.44+14x+2y=10,36\\ BTE:6x+2y=0,55.4\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,36\\ y=0,02\end{array}\right.\end{array}$

$\begin{array}{l}\Rightarrow E\text{\hspace{0.17em}\hspace{0.17em}gồ}m\left\{\begin{array}{l}Y:0,02mol\\ X:0,12-0,02.2=0,08mol\end{array}\right.\Rightarrow \overline{C}=\frac{0,48}{0,1}=4,8\Rightarrow X\text{\hspace{0.17em}\hspace{0.17em}hoặ}cY\text{\hspace{0.17em}\hspace{0.17em}có\hspace{0.17em}\hspace{0.17em}}4C.\\ •\text{\hspace{0.17em}\hspace{0.17em}Nế}uXcó4C\left(C_2{H}_{3}COOC{H}_3\right)\Rightarrow C_Y=\frac{0,48-0,08.4}{0,02}=8\Rightarrow Y\mathrm{là}C{H}_{3}OOC{C}_4{H}_{8}COOC{H}_3\Rightarrow \%Y=33,6\%.\\ •NếuY\text{\hspace{0.17em}c\hspace{0.17em}ó\hspace{0.17em}\hspace{0.17em}}4C\left(C{H}_{3}OOCCOOC{H}_3\right)\Rightarrow C_Y=\frac{0,48-0,02.4}{0,08}=5\Rightarrow X\text{\hspace{0.17em}là\hspace{0.17em}\hspace{0.17em}}{C}_3{H}_{5}COOC{H}_3\Rightarrow \%Y=22,78\%.\\ \Rightarrow Chọn\boxed{\%Y=33,6\%}\end{array}$

Chọn A