Phần $2:n_{Ag}=0,09 mol>\frac{0,08}{2} mol\Rightarrow ancol:CH_3OH$$*)$Phần $1:$
$CH_3OH\overset{+[O]}{\rightarrow}HCOOH+H_2O$
$x x x$
$CH_3OH\overset{+[O]}{\rightarrow}HCHO+H_2O$
$y y y$
$hh+Na\Rightarrow n_{H_2}=\frac{0,04+x}{2}=\frac{0,504}{22,4}\Rightarrow x=0,005 mol$
$*)$Phần $2$
$hh+AgNO_3/NH_3\Rightarrow 2x+4y=\frac{9,72}{108}\Rightarrow y=0,02 mol$
$\Rightarrow \%CH_3OH=\frac{(x+y).100}{0,04}=62,5\%$