Đáp án:

 b) 3,36 l

c) 2,67 g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
b)\\
nKCl{O_3} = \frac{{12,25}}{{122,5}} = 0,1\,mol\\
 =  > n{O_2} = 0,15\,mol\\
V{O_2} = 0,15 \times 22,4 = 3,36l\\
c)\\
nFe = \frac{{5,6}}{{56}} = 0,1\,mol\\
3Fe + 2{O_2} \to F{e_3}{O_4}\\
\frac{{0,1}}{3} < \frac{{0,15}}{2}
\end{array}\)

=> O2 dư

\(\begin{array}{l}
n{O_2} = 0,15 - \frac{1}{{15}} = \frac{1}{{12}}(mol)\\
m{O_2} = \frac{{32}}{{12}} = 2,67g
\end{array}\)

$n_{KClO3}=12,25/122,5=0,1mol$

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

$b/$

$\text{Theo pt :}$

$n_{O_2}=3/2.n_{KClO3}=3/2.0,1=0,15mol$

$⇒a=0,15.22,4=3,36l$

$c/$

$n_{Fe}=5,6/56=0,1mol$

                               $3Fe  +  2O2\overset{t^o}\to Fe3O4$

$\text{Theo pt :     3 mol    2 mol}$

$\text{Theo đbài :  0,1mol  0,15mol}$

$\text{⇒Sau pư O2 dư 1/12 mol}$

$⇒m_{O_2 dư}=1/12.32=8/3g$