X; Y; Z là ba este đều mạch hở và không chứa nhóm chức khác (trong đó X; Y đều

* Hướng dẫn giải

$$\begin{gathered} \mathbf{Đáp}\mathbf{ }\mathbf{án}\mathbf{ }\mathbf{đúng}\mathbf{ }\mathbf{là}\mathbf{:}\mathbf{ }\mathbf{C} \\ \{\begin{array}{l}\mathrm{X}\\ \mathrm{Y}\\ \mathrm{Z}\end{array}\underset{}{\overset{+\mathrm{NaOH}}{\to }}|\begin{array}{l}\text{F}\underset{}{\overset{+{\mathrm{O}}_2}{\to }}\left\{\begin{array}{l}{\mathrm{Na}}_2{\mathrm{CO}}_3\text{ 0,26}\\ {\mathrm{CO}}_2\\ {\mathrm{H}}_2\mathrm{O}\end{array}\right.\\ \mathrm{Ancol}\end{array} \\ \mathrm{Ancol}\underset{}{\overset{+\mathrm{Na}}{\to }} \mathrm{khối} \mathrm{lượng} \mathrm{bình} \mathrm{tăng} 16,2 \\ \to \{\begin{array}{l}{\text{n}}_{\mathrm{OH}\left(\mathrm{ancol}\right)}=0,26.2=0,52\\ {\mathrm{m}}_{\mathrm{ancol}}=16,2+0,52=16,72\end{array}\underset{}{\overset{}{\to }}0,26<{\mathrm{n}}_{\mathrm{ancol}}<0,52 \\ \underset{}{\overset{}{\to }}32,2<{\mathrm{M}}_{\mathrm{ancol}}<64,4 \end{gathered}$$

$$\begin{gathered} \to \mathrm{Ancol}\{\begin{array}{l}{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\text{ a}\\ {\mathrm{C}}_2{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2 \text{ b}\end{array}\underset{}{\overset{}{\to }}\{\begin{array}{l}\mathrm{a}+2\mathrm{b}=0,52\\ 46\mathrm{a}+62\mathrm{b}=16,72\end{array}\underset{}{\overset{}{\to }}\{\begin{array}{l}\mathrm{a}=0,04\\ \mathrm{b}=0,24\end{array} \\ \mathrm{BTKL}:{\mathrm{m}}_{\mathrm{F}}=38,56+0,52.40-16,72=42,64 \\ \text{F}\{\begin{array}{l}{\mathrm{R}}_1\mathrm{COONa}\text{ 0,26}\\ {\mathrm{R}}_2\mathrm{COONa}\text{ 0,26}\end{array}\underset{}{\overset{}{\to }}{\mathrm{m}}_{\mathrm{F}}=0,26.({\mathrm{R}}_1+{\mathrm{R}}_2)+0,52.67=42,64 \\ \underset{}{\overset{}{\to }}{\mathrm{R}}_1+{\mathrm{R}}_2=30\underset{}{\overset{}{\to }}\{\begin{array}{l}{\mathrm{R}}_1=1\\ {\mathrm{R}}_2=29\end{array} \\ \underset{}{\overset{}{\to }}\{\begin{array}{l}\mathrm{X}:{\mathrm{HCOOC}}_2{\mathrm{H}}_5\text{ 0,02}\underset{}{\overset{}{\to }}\mathrm{\%}{\mathrm{m}}_{{\mathrm{HCOOC}}_2{\mathrm{H}}_5}=3,84\mathrm{\%}\\ \mathrm{Y}:{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{COOC}}_2{\mathrm{H}}_5\text{ 0,02}\\ \mathrm{Z}:{\mathrm{HCOOCH}}_2-{\mathrm{CH}}_2-{\mathrm{OOCC}}_2{\mathrm{H}}_5\text{ 0,24}\end{array} \end{gathered}$$