2/
$PTPƯ:$
$2C_2H_5OH+2Na\xrightarrow{} 2C_2H_5ONa+H_2↑$ $(1)$
$2CH_3OH+2Na\xrightarrow{} 2CH_3ONa+H_2↑$ $(2)$
$n_{H_2}=\dfrac{5,6}{22,4}=0,25mol.$
$Theo$ $pt1:$ $n_{H_2}=2\dfrac{1}{2}n_{C_2H_5OH}$
$Theo$ $pt2:$ $n_{H_2}=\dfrac{1}{2}n_{CH_3OH}$
Gọi $n_{C_2H_5OH}$ là a (mol), $n_{CH_3OH}$ là b (mol)
Theo đề bài ta có hê pt: $\left \{ {{46a+32b=18,8} \atop {0,5a+0,5b=0,25}} \right.$ $⇒\left \{ {{a=0,2} \atop {b=0,3}} \right.$
$⇒\%m_{C_2H_5OH}=\dfrac{0,2.46}{18,8}.100\%=48,94\%$
$⇒\%m_{CH_3OH}=\dfrac{0,3.32}{18,8}.100\%=51,06\%$
chúc bạn học tốt!