Đáp án:

 b) 8,4l

c) 30,625 g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Zn + {O_2} \to 2ZnO\\
b)\\
nZn = \dfrac{m}{M} = \dfrac{{48,6}}{{65}} \approx 0,75\,mol\\
n{O_2} = \dfrac{{0,75}}{2} = 0,375\,mol\\
V{O_2} = n \times V = 0,375 \times 22,4 = 8,4l\\
c)\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
nKCl{O_3} = \dfrac{{0,375 \times 2}}{3} = 0,25\,mol\\
mKCl{O_3} = n \times M = 0,25 \times 122,5 = 30,625g
\end{array}\)

$n_{Zn}=48,6/65≈0,75mol$

$a/2Zn+O_2\overset{t^o}\to 2ZnO$

b/Theo pt :

$n_{O_2}=1/2.n_{Zn}=1/2.0,75=0,375mol$

$⇒V_{O_2}=0,375.22,4=8,4l$

$c/2KClO_3\overset{t^o}\to 2KCl+3O_2$

Theo pt :

$n_{KClO_3}=2/3.n_{O_2}=2/3.0,375=0,25mol$

$⇒m_{KClO_3}=0,25.122,5=30,625g$