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