X, Y là hai axit cacboxylic đều mạch hở Z là ancol no

* Hướng dẫn giải

Đáp án D

Ancol + Na:

C_nH_2n+2O₂ + Na → C_nH_2nO₂Na₂ + H₂

0,13                           ←            0,13

m _{bình tăng} = m _ancol – m_Na => m _ancol = 9,62 + 0,13.2 = 9,88 (g)

M _ancol = 9,88 : 0,13 = 76 (C₃H₈O₂)

$\stackrel{BTKL}{\to }{m}_F =m_E+m_{NaOH}-m_{ancol}=19,43 +0,2.22,4-9,88=17,55\left(g\right)$

$$\begin{gathered} \stackrel{BTNTO}{\to } n_{{}_{C{O}_2}}=\frac{2 n_{muoi}+2n_{O_2}-{n_H}_{2^O}-3n_{N{a}_{2}C{O}_3}}{2}=0,3\left(mol\right) \\ \stackrel{BTKL}{\to } m_{muoi}=0,3.44+0,2.18+0,1.106-0,35.32=16,2\left(g\right) \\ {m}_{{H_2}^{O\left(F\right)}}=17,55-16,2=1,35\left(g\right) \\ \to n_{axit}=n_{{H_2}^{O\left(F\right)}}=0,075\left(mol\right) \\ \to n_T=\frac{n_{NaOH-}{n}_{axit}}{2}=0,0625\left(mol\right) \\ \to n_X=n_{ancol}- n_T=0,13-0,0625=0,0675\left(mol\right) \\ \to \%m_z=\frac{0,0675.76}{19,43}.100\%=26,40\% \end{gathered}$$