(Lưu ý: Câu b là 44,8 lít khí oxi nha bạn)
$PTPƯ:2KClO_3\xrightarrow{t^o} 2KCl+3O_2$
$a,n_{O_2}=\dfrac{48}{32}=1,5mol.$
$Theo$ $pt:$ $n_{KClO_3}=\dfrac{2}{3}n_{O_2}=1mol.$
$⇒m_{KClO_3}=1.122,5=122,5g.$
$b,n_{O_2}=\dfrac{44,8}{22,4}=2mol.$
$Theo$ $pt:$ $n_{O_2}=\dfrac{2}{3}n_{O_2}=\dfrac{4}{3}mol.$
$⇒m_{KClO_3}=\dfrac{4}{3}.122,5=163,33g.$
chúc bạn học tốt!