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