Em tham khảo nha:

\(\begin{array}{l}
4)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
{n_{{C_2}{H_5}OH}} = \dfrac{{4,6}}{{46}} = 0,1\,mol\\
{n_{{H_2}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{V_{{H_2}}} = 0,05 \times 22,4 = 1,12l\\
6)\\
a)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}\\
{n_{{H_2}}} = \dfrac{{1,792}}{{22,4}} = 0,08\,mol\\
hh:{C_2}{H_5}OH(a\,mol),{C_6}{H_5}OH(b\,mol)\\
\left\{ \begin{array}{l}
46a + 94b = 11,2\\
0,5a + 0,5b = 0,08
\end{array} \right.\\
 \Rightarrow a = b = 0,08\,mol\\
{m_{{C_2}{H_5}OH}} = 0,08 \times 46 = 3,68g\\
{m_{{C_6}{H_5}OH}} = 11,2 - 3,68 = 7,52g\\
b)\\
\% {m_{{C_2}{H_5}OH}} = \dfrac{{3,68}}{{11,2}} \times 100\%  = 32,86\% \\
\% {m_{{C_6}{H_5}OH}} = 100 - 32,86 = 67,14\% 
\end{array}\)

Giải thích các bước giải:

 4.

$C_2H_5OH+Na\to C_2H_5ONa + \dfrac 12H_2\\ n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1\ \text{mol}\\\to n_{H_2}=0,1.\dfrac 12=0,05\ \text{mol}\to V_{H_2}=0,05.22,4=1,12\ text{l}$

6. 

a.

Gọi số mol phenol, ancol etylic lần lượt là a, b

$\to 94a+46b=11,2$

PTHH: $C_6H_5OH+Na\to C_6H_5ONa + \dfrac 12H_2\\ C_2H_5OH+Na\to C_2H_5OH+\dfrac 12H_2\\\to n_{H_2}=\dfrac 12a+\dfrac 12b =\dfrac{1,792}{22,4}\to a+b=0,16\ \text{mol}$

$\to a=0,08; b=0,08$

$\to m_{phenol}=0,08.94=7,52\ \text{gam}\to m_{C_2H_5OH}=11,2-7,52=3,68\ \text{gam}$

b. $\%m_{phenol}=\dfrac{7,52}{11,2}.100\%=67,14\%\to\%m_{C_2H_5OH}=100\%-67,14\%=32,86\%$