Đốt cháy hoàn toàn a gam triglixer...

B

- Hướng dẫn giải

Đáp án B

$$\begin{gathered} \left\{\begin{array}{l}BTNT O: 6\underset{?}{\underbrace{n_{X \left(C_3{H}_5{\left(OOC\overline{R}\right)}_3\right)}}}+2\underset{3,26}{\underbrace{n_{O_2}}}=2\underset{2,28}{\underbrace{n_{C{O}_2}}}+\underset{2,2}{\underbrace{n_{H_{2}O}}}\\ m_X=m_C+m_H+m_{O/X}\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}{n}_{C_3{H}_5{\left(OOC\overline{R}\right)}_3}=0,04; n_{O/C_3{H}_5{\left(OOC\overline{R}\right)}_3}=0,04.6=0,24\\ m_X=2,28.12+2,2.2+0,24.16=35,6\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}{n}_{NaOH pư}=3n_{C_3{H}_5{\left(OOC\overline{R}\right)}_3}=0,04.3=0,12\\ n_{C_3{H}_5{\left(OH\right)}_3}=n_{C_3{H}_5{\left(OOC\overline{R}\right)}_3}=0,04\end{array}\right. \\ \Rightarrow m_{muối}=\underset{35,6}{\underbrace{m_X}}+\underset{0,12.40}{\underbrace{m_{NaOH}}}-\underset{0,04.92}{\underbrace{m_{C_3{H}_5{\left(OH\right)}_3}}}=\boxed{36,72 gam} \end{gathered}$$