Câu $1:$$*)n_{CO_2}=n_{H_2O}\Rightarrow n_{C_2H_5OH}=n_{CH_2=CH-CHO}$
$X: \begin{cases}C_2H_5OH:x mol \\ CH_2=CH-CHO:x mol\\CH_3-CHO:y mol \end{cases}$
$C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O$
$x 3x$
$CH_2=CH-CHO+3,5O_2\rightarrow 3CO_2+2H_2O$
$x 3,5x$
$CH_3-CHO+2,5O_2\rightarrow 2CO_2+2H_2O$
$y 2,5y$
$\Rightarrow \begin{cases}m=102x+44y \\ 1,9368m=32(6,5x+2,5y) \end{cases}\Rightarrow 102x+44y=\frac{32(6,5x+2,5y)}{1,9368}$
$\Rightarrow y\approx 2x\Rightarrow \%m_{CH_3-CHO}=\frac{44\times (2x)\times 100}{102x+44\times (2x)}=46,3\%$