Cho m gam một axit cacboxylic X...

B

- Hướng dẫn giải

Đáp án B

$$\begin{gathered} + \mathrm{Bản} \mathrm{chất} \mathrm{phản} \mathrm{ứng}: \left\{\begin{array}{l}-\mathrm{COOH}+{\mathrm{NaHCO}}_3\stackrel{}{\to }-\mathrm{COONa}+{\mathrm{CO}}_2\uparrow +{\mathrm{H}}_2\mathrm{O}\\ 2(-\mathrm{COOH})+\mathrm{Ca}(\mathrm{OH}{)}_2\stackrel{}{\to }(-\mathrm{COO}{)}_2\mathrm{Ca}+2{\mathrm{H}}_2\mathrm{O}\end{array}\right. \\ + \left\{\begin{array}{l}{\mathrm{n}}_{-\mathrm{C}\text{OO}\mathrm{H}}={\mathrm{n}}_{{\mathrm{NaHCO}}_3}=0,1\\ {\mathrm{n}}_{{(-\mathrm{COO})}_2\mathrm{Ca}}=0,5{\mathrm{n}}_{\mathrm{COOH}}=0,05\end{array}\right.\Rightarrow {\mathrm{M}}_{{(-\mathrm{COO})}_2\mathrm{Ca}}=\frac{9,1}{0,05}=182 \\ \bullet \mathrm{Nếu} \mathrm{muối} \mathrm{có} \mathrm{dạng} \mathrm{R}(\mathrm{COO}{)}_2\mathrm{Ca}\Rightarrow \mathrm{R}=54 \left(\mathrm{loại}\right). \\ \bullet \mathrm{Nếu} \mathrm{muối} \mathrm{có} \mathrm{dạng} (\mathrm{RCOO}{)}_2\mathrm{Ca}\Rightarrow \mathrm{R}=27\Rightarrow \boxed{{\mathrm{CH}}_2=\mathrm{CHCOOH} (\mathrm{axit} \mathrm{acrylic})} \end{gathered}$$