Đốt cháy hoàn toàn m gam một este đơn chức X cần vừa hết 10,08 lít oxi (đktc),

* Hướng dẫn giải

$\begin{array}{l}•X\text{\hspace{0.17em}\hspace{0.17em}kh}ôngphailaestecuaphenol\\ +\left\{\begin{array}{l}{n}_{X\left(C_x{H}_y{O}_2\right)}=n_{NaOH}=\mathrm{0,1}\\ 2\underset{\mathrm{0,1}}{\underbrace{n_{C_x{H}_y{O}_2}}}+2\underset{\mathrm{0,45}}{\underbrace{n_{O_2}}}=2\underset{\mathrm{0,4}}{\underbrace{n_{C{O}_2}}}+\underset{?}{\underbrace{n_{H_{2}O}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{H_{2}O}=\mathrm{0,3}\\ n_C:n_H:n_O=4:6:2\\ X\text{\hspace{0.17em}\hspace{0.17em}là\hspace{0.17em}\hspace{0.17em}}{C}_4{H}_6{O}_2\end{array}\right.\end{array}$

$\begin{array}{l}+X\mathit{ }có\mathit{ 3 }đồng\mathit{ }phân\mathit{ }cấu\mathit{ }tạo\mathit{ }có\mathit{ }dạng{C}_x{H}_{y}OOCH\text{\hspace{0.17em}\hspace{0.17em}là}:\hspace{0.17em}\\ \boxed{HCOOCH=CH-C{H}_3;HCOOC\left(C{H}_3\right)=C{H}_2;HCOOC{H}_{2}CH=C{H}_2}\\ •X là este của phenol\\ +\left\{\begin{array}{l}{n}_{X\left(C_x{H}_y{O}_2\right)}=\mathrm{0,5}{n}_{NaOH}=\mathrm{0,05}\\ 2\underset{\mathrm{0,05}}{\underbrace{n_{C_x{H}_y{O}_2}}}+2\underset{\mathrm{0,45}}{\underbrace{n_{O_2}}}=2\underset{\mathrm{0,4}}{\underbrace{n_{C{O}_2}}}+\underset{?}{\underbrace{n_{H_{2}O}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{H_{2}O}=\mathrm{0,2}\\ n_C:n_H:n_O=8:8:2\\ X\hspace{0.17em}\mathrm{là}{C}_8{H}_8{O}_2\end{array}\right.\\ +X có 3 đồng phân:\end{array}$