Theo bài ra ta có công thức ancol $X:C_nH_{2n+1}CH_2OH$ $\Rightarrow Y:\begin{cases}C_nH_{2n+1}CH_2OH:3x mol \\ C_nH_{2n+1}CHO:3y mol\\C_nH_{2n+1}COOH:3z mol \end{cases}$
$*)P_1:$
$n_{Ag}=0,2 mol\Rightarrow n_{C_nH_{2n+1}CHO}=0,1 mol\Rightarrow y=0,1 mol$
$*)P_2:$
$n_{CO_2}=0,1 mol\Rightarrow n_{C_nH_{2n+1}COOH}=0,1 mol\Rightarrow z=0,1 mol$
$*)P_3:$
$C_nH_{2n+1}CH_2OH+Na\rightarrow C_nH_{2n+1}CH_2ONa+0,5H_2\uparrow$
$x x 0,5x$
$C_nH_{2n+1}COOH+Na\rightarrow C_nH_{2n+1}COONa+0,5H_2\uparrow$
$z z 0,5z$
$\Rightarrow \begin{cases}0,5x+0,5z=\frac{4,48}{22,4} \\y=0,1\\ z=0,1\\x(14n+54)+z(14n+68)=28,6 \end{cases}\Rightarrow \begin{cases}x=0,3 mol \\ y=0,1 mol\\z=0,1 mol\\n=1 \end{cases}$
$\Rightarrow X:CH_3CH_2OH:etanol$
$\Rightarrow \%X=\frac{(3y+3z)\times 100}{(3x+3y+3z)}=40\%$