$n_{KClO3}=245/122,5 = 2 mol$
$a/$
$PTHH :$
$2KClO3\overset{t^o}\to 2KCl + 3O2$
$\text{Theo pt :}$
$n_{KCl}=n_{KClO3}=2mol$
$⇒m=2,74,5=148g$
$n_{O_2}=3/2.n_{KClO3}=3/2.2=3mol$
$⇒V_{O_2}= 3.22,4 = 67,2l$
$b/$
$n_{Mg}=24/24= 1 mol$
$n_{O_2}=67,2/22,4 = 3 mol$
$PTHH :$
$2Mg + O2\overset{t^o}\to 2MgO$
$\text{Ta có tỷ lệ :}$
$\dfrac{1}{2} <\dfrac{3}{1}$
$\text{⇒Sau pư O2 dư}$
$\text{Theo pt :}$
$⇒n_{MgO}= n_{Mg} = 1 mol$
$⇒m_{MgO}= 1.40=40g$