Thủy phân hoàn toàn 10,75 gam este X (có công thức phân tử dạng CnH2n-2O2

* Hướng dẫn giải

$$\begin{gathered} Đáp án B \\ + TH1: X là \underset{x mol}{\underbrace{HCOOCH=CHR}} \\ \stackrel{NaOH}{\to }\left\{\begin{array}{l}\underset{x mol}{\underbrace{HCOONa}}\\ \underset{x mol}{\underbrace{RC{H}_{2}CHO}}\end{array}\right. \\ \stackrel{AgN{O}_3/N{H}_3}{\to }\underset{4x}{\underbrace{Ag}} \\ \Rightarrow \left\{\begin{array}{l}x=\frac{10,75}{14n+30}\Rightarrow \left\{\begin{array}{l}x=0,125\\ n=4\end{array}\right.\\ 4x = 0,5\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}X: HCOOCH=CHC{H}_3\\ gồm 2 đp cis và trans\end{array}\right. \\ TH2: X là \left\{\begin{array}{l}\underset{x mol}{\underbrace{HCOOR}}\\ \underset{x mol}{\underbrace{HCOOCH=CHR}}\end{array}\right.\stackrel{NaOH}{\to } \\ \left\{\begin{array}{l}\underset{x mol}{\underbrace{HCOOR}}\stackrel{AgN{O}_3/N{H}_3}{\to }\underset{2x}{\underbrace{Ag}}\\ \underset{x mol}{\underbrace{RC{H}_{2}CHO}}\stackrel{AgN{O}_3/N{H}_3}{\to }\underset{2x}{\underbrace{Ag}}\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}x=\frac{10,75}{14n+30}\Rightarrow \left\{\begin{array}{l}x=0,25\\ n=0,928\left(loại\right)\end{array}\right.\\ 2x = 0,5\end{array}\right. \end{gathered}$$