Cho hỗn hợp E gồm ba chất X, Y và ancol propylic....

A

- Hướng dẫn giải

\(\begin{array}{l}
\left\{ \begin{array}{l}
{C_3}{H_7}OH\\
{C_{\bar n}}{H_{2\bar n{\rm{ }} + {\rm{ }}4{\rm{ }} - {\rm{ }}2k}}{N_2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{C_3}{H_6}{H_2}O\\
{C_{\bar n}}{H_{2\bar n{\rm{ }} + {\rm{ }}4{\rm{ }} - {\rm{ }}2k}}{N_2}
\end{array} \right.\left\{ \begin{array}{l}
{{\rm{n}}_{{O_2}}}{\rm{ =  1,5nC}}{{\rm{O}}_2}\\
{{\rm{n}}_{{O_2}}}{\rm{ =  0,67  >  0,42}}{\rm{.1,5}} \to {\rm{4  -  2k  >  0 }} \to {\rm{k  =  1 }}
\end{array} \right.\\
{\rm{0,1 (mol) E}}\left\{ \begin{array}{l}
{C_3}{H_7}OH{\rm{: x}}\\
{C_{\bar n}}{H_{2\bar n{\rm{ }} + {\rm{ 2 }}}}{N_2}:{\rm{ y}}
\end{array} \right. \to \left\{ \begin{array}{l}
C{O_2}:{\rm{ 0,42}}\\
{H_2}O\\
{N_2}
\end{array} \right. \to \left\{ \begin{array}{l}
{n_E}:{\rm{ }}x + y = 0,1\\
BT(e):{\rm{ 18x  +  y(6\bar n  +  2)  =  2,68}}\\
{{\rm{n}}_{C{O_2}}}{\rm{: 3x  +  \bar ny  =  0,42 }}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + y = 0,1\\
{\rm{6(3x  +  y\bar n)  +  2y  =  2,68}}\\
{\rm{3x  +  \bar ny  =  0,42 }}
\end{array} \right. \to \left\{ \begin{array}{l}
y = 0,08\\
x = 0,02\\
{\rm{\bar n  =  4,5}}
\end{array} \right. \to \left\{ \begin{array}{l}
{C_4}{H_{10}}{N_2}:0,04\\
{C_5}{H_{12}}{N_2}:0,04
\end{array} \right.\\
 \to \% Y = \frac{{0,04*100}}{{0,04*100 + 0,04*86 + 0,02*60}}*100\%  = 46,3\% 
\end{array}\)