a) PTHH : 2KClO3 -> 2KCl + 3O2 

 nKClO3: 245/122,5 = 2 mol
cứ 2 mol KClO3 -> 2 mol KCl -> 3 mol O2
      2 mol              -> 2 mol         -> 3 mol
-> m = 2*74,5=149(g)
V = 3*22,4 = 67,2(l)
b) PTHH : 2Mg + O2 -> 2MgO
nMg: 24/24= 1 mol
nO2: 67,2/22,4 = 3 mol
ta có: nMg/2 < nO2/1 nên O2 dư -> tính theo Mg
->nMgO : 1*2/2 = 1 mol
->mMgO = 1*40=40(g)

$n_{KClO3}=245/122,5 = 2 mol$

$a/$

$PTHH :$

$2KClO3\overset{t^o}\to 2KCl + 3O2$

$\text{Theo pt :}$

$n_{KCl}=n_{KClO3}=2mol$

$⇒m=2,74,5=148g$
$n_{O_2}=3/2.n_{KClO3}=3/2.2=3mol$
$⇒V_{O_2}= 3.22,4 = 67,2l$
$b/$

$n_{Mg}=24/24= 1 mol$
$n_{O_2}=67,2/22,4 = 3 mol$

$PTHH :$

$2Mg + O2\overset{t^o}\to 2MgO$
$\text{Ta có tỷ lệ :}$

$\dfrac{1}{2} <\dfrac{3}{1}$

$\text{⇒Sau pư O2 dư}$

$\text{Theo pt :}$
$⇒n_{MgO}= n_{Mg} = 1 mol$
$⇒m_{MgO}= 1.40=40g$