$\begin{cases}n_{KClO_3}=0,2 mol \\n_{O_2}=0,18 mol \end{cases}\Rightarrow m_X=24,5-0,18.32=18,74 gam$$\Rightarrow n_{KCl}= \frac{18,74.50,69}{74,5.100}=0,1275 mol$
$2KClO_3\rightarrow 2KCl+3O_2$
$x x 1,5x$
$4KClO_3\rightarrow KCl+3KClO_4$
$4y y 3y$
$\Rightarrow 1,5x=0,12 mol\Rightarrow y=0,1275-0,12=0,0075 mol$
$n_{KClO_3 con lai}=0,2-(x+4y)=0,05 mol$
$\%KClO_3 trong X=\frac{0,05.122,5.100}{18,74}=32,68\%$