A. 50,84%.
B. 61,34%.
C. 69,53%.
D. 53,28%.
C
\(\left\{ \begin{array}{l}
BTE:0,45.6 - 10a - 22b = 0,5.4\\
\frac{{{n_E}}}{{{n_{COO}}}} = \frac{{a + b}}{{2a + 3b}} = \frac{{0,16}}{{0,42}}
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,015\\
b = 0,025
\end{array} \right.\)
\(\begin{array}{l} {\rm{0,015n + 0,025m = 0,45}} \to \left\{ \begin{array}{l} 3n + 5m = 90\\ n > 5;m \ge 12 \end{array} \right. \to \left\{ \begin{array}{l} n = 10\\ m = 12{\rm{ (}}{{\rm{C}}_2}{H_3}COO{)_3}{C_3}{H_5} \end{array} \right.\\ {\rm{0,16 mol E}}\left\{ \begin{array}{l} {C_{10}}{H_{18}}{O_4}{\rm{ 0,06}}\\ {{\rm{C}}_{12}}{H_{14}}{O_6}{\rm{ 0,1}} \end{array} \right. \to \left\{ \begin{array}{l} RCOONa\\ {C_2}{H_3}COONa{\rm{ 0,3}} \end{array} \right. + \left\{ \begin{array}{l} {C_3}{H_6}{(OH)_2}{\rm{ 0,06}}\\ {{\rm{C}}_3}{H_5}{(OH)_3}{\rm{ 0,1}} \end{array} \right.\\ BTKL:{m_{muoi}} = a = (0,06.202 + 0,1.254 + 0,42.40) - (0,06.76 + 0,1.92) = 40,56g\\ \% {m_{{C_2}{H_3}COONa}} = \frac{{0,3.94}}{{40,56}}.100 = 69,53\% \end{array}\)
Câu hỏi trên thuộc đề trắc nghiệm dưới đây !
Copyright © 2021 HOCTAP247