Gọi \(\left\{{}\begin{matrix}n_{C_2H_5OH}=a\left(mol\right)\\n_{C_6H_5OH}=b\left(mol\right)\end{matrix}\right.;n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}46a+94b=14\\a+b=0,1.2\end{matrix}\right.\Leftrightarrow a=b=0,1\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{0,1.46}{0,1.\left(46+94\right)}.100\%=32,86\%\\\%m_{C_6H_5OH}=100\%-32,86\%=67,14\%\end{matrix}\right.\)