\(\begin{array}{l}
Dat\,n_{CH_3OH}=x(mol);n_{C_2H_5OH}=y(mol)\\
\to 32x+46y=15,6(1)\\
n_{H_2}=\frac{4,48}{22,4}=0,2(mol)\\
2CH_3OH+2Na\to 2CH_3ONa+H_2\\
2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2\\
Theo\,PT:\,x+y=2n_{H_2}=0,4(2)\\
(1)(2)\to x=y=0,2(mol)\\
\to \%n_{CH_3OH}=\%n_{C_2H_5OH}=\frac{0,2}{0,2+0,2}.100\%=50\%
\end{array}\)

Đáp án + Giải thích các bước giải:

Gọi $x\text{ mol} = n_{CH_3OH}, y\text{ mol} = n_{C_2H_5OH}$
$m_{CH_3OH} + m_{C_2H_5OH} = 15,6$
$\Rightarrow 32x + 46y = 15,6(1)$

 PTHH:
$2CH_3OH + Na \longrightarrow 2CH_3ONa + H_2 \uparrow$

$2C_2H_5OH + 2Na \longrightarrow 2C_2H_5OH + H_2 \uparrow$

$n_{H_2} = \dfrac{V}{22,4} = \dfrac{4,48}{22,4} = 0,2\text{ mol}$

Theo PTHH:
$n_{H_2} = \dfrac{1}{2}n_{CH_3OH} + \dfrac{1}{2}n_{C_2H_5OH} = 0,2\text{ mol}$
$\Rightarrow \dfrac{1}{2}x + \dfrac{1}{2}y = 0,2(2)$

Từ $(1)$ và $(2)$, ta có:
$\begin {cases} 32x + 46y = 15,6 \\ \dfrac{1}{2}x + \dfrac{1}{2}y = 0,2 \end {cases}$
$\Leftrightarrow \begin {cases} x = 0,2 \\ y = 0,2 \end {cases}$
$\Rightarrow \%n_{CH_3OH} = \dfrac{0,2}{0,2 + 0,2} = 50\%$

$\%n_{C_2H_5OH} = 100 - 50\% = 50\%$