Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
hh:{C_6}{H_5}OH(a\,mol),{C_2}{H_5}OH(b\,mol)\\
\left\{ \begin{array}{l}
0,5a + 0,5b = 0,25\\
94a + 46b = 32,6
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,3\\
\% {m_{{C_6}{H_5}OH}} = \dfrac{{0,2 \times 94}}{{32,6}} \times 100\%  = 57,67\% \\
\% {m_{{C_2}{H_5}OH}} = 100 - 57,67 = 42,33\% \\
2)\\
a)\\
2{C_3}{H_7}OH + 2Na \to 2{C_3}{H_7}ONa + {H_2}\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{7,84}}{{22,4}} = 0,35\,mol\\
hh:{C_3}{H_7}OH(a\,mol),{C_2}{H_5}OH(b\,mol)\\
\left\{ \begin{array}{l}
0,5a + 0,5b = 0,35\\
60a + 46b = 36,4
\end{array} \right.\\
 \Rightarrow a = 0,3;b = 0,4\\
\% {m_{{C_3}{H_7}OH}} = \dfrac{{0,3 \times 60}}{{35,4}} \times 100\%  = 50,85\% \\
\% {m_{{C_2}{H_5}OH}} = 100 - 50,85 = 49,15\% 
\end{array}\)