$n_{KCl}=n_{KClO_3}=\frac{14,9}{74,5}= 0,2mol$
$\Rightarrow m_{KClO_3}=0,2.122,5= 24,5g$
$m_Y= 14,9:27,7261\%= 53,74g$
$n_{O_2}= \frac{8,288}{22,4}=0,37 mol$
Sơ đồ: $X \to O_2+ Y$
BTKL, $m_X=53,74+0,37.32= 65,58g$
$\Rightarrow m_{KMnO_4}=65,58-24,5=41,08g$