Cho 7,0 gam hỗn hợp A gồm Phenol (C6H5OH) và Etanol (C2H5OH)  tác dụng với Natri (dư) thu được 1,12 lít khí hiđro H2 (ở đk

* Hướng dẫn giải

\(\begin{array}{l}
{n_{{H_2}}} = \frac{{1,12}}{{22,4}} = 0,05mol\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa\,\,\,\,\,\, + \,\,\,\,{H_2} \uparrow \\
\,\,\,xmol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{x}{2}mol\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa\,\,\,\,\,\, + \,\,\,\,{H_2} \uparrow \\
\,\,\,ymol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{y}{2}mol\\
\left\{ \begin{array}{l}
94x + 46y = 7\\
\frac{x}{2} + \frac{y}{2} = 0,05
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 0,05mol\\
y = 0,05mol
\end{array} \right.\\
\% {m_{{C_6}{H_5}OH}} = \frac{{94x}}{7}100 = 67,14\% \\
\% {m_{{C_2}{H_5}OH}} = \frac{{46y}}{7}100 = 32,86\% 
\end{array}\)