a,
$n_{H_2}=0,3 mol$
Gọi chung 2 ancol là $C_nH_{2n+2}O$ (0,3.2=0,6 mol)
$2C_nH_{2n+1}OH+2Na\to 2C_nH_{2n+1}ONa+ H_2$ (*)
$\Rightarrow \overline{M}=\frac{22}{0,6}=36,67=14n+18$
$\Leftrightarrow n=1,3$
Vậy 2 ancol là $CH_3OH$ (a mol), $C_2H_5OH$ (b mol)
b,
Ta có $32a+46b=22$ (1)
Theo (*): $0,5a+0,5b=0,3$ (2)
(1)(2) $\Rightarrow a=0,4; b=0,2$
$\%m_{CH_3OH}=\frac{0,4.32.100}{22}= 58,2\%$
$\%m_{C_2H_5OH}= 41,8\%$
c,
$n_{\text{ancol}}=n_{\text{muối}}$ (bảo toàn C)
$\Rightarrow m_{\text{muối}}= 0,4.54+0,2.68=35,2g$
d,
$CH_3OH\to HCHO \to 4Ag$
$C_2H_5OH \to CH_3CHO \to 2Ag$
$\Rightarrow n_{Ag}= 4a+2b=2mol$
$a=2.108=216g$
