Em tham khảo nha :
\(\begin{array}{l}
1)\\
a)\\
hh:Fe(a\,mol),Zn(b\,mol)\\
2a - b = 0(1)\\
56a + 65b = 18,6(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,1;b = 0,2\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
2Zn + {O_2} \xrightarrow{t^0} 2ZnO\\
{n_{{O_2}}} = \dfrac{2}{3}{n_{Fe}} + \dfrac{{{n_{Zn}}}}{2} = \frac{1}{6}mol\\
{V_{{O_2}}} = n \times 22,4 = \dfrac{1}{6} \times 22,4 = 3,73l\\
b)\\
{V_{kk}} = 5{V_{{O_2}}} = 5 \times 3,73 = 18,65l\\
2)\\
a)\\
2KMn{O_4} \xrightarrow{t^0} {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
b)\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \frac{{5,6}}{{22,4}} = 0,25mol\\
hh:KMn{O_4}(a\,mol),KCl{O_3}(b\,mol)\\
\dfrac{a}{2} + \dfrac{3}{2}b = 0,25(1)\\
158a + 122,5b = 43,35(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,2;b = 0,1\\
{m_{KMn{O_4}}} = n \times M = 0,2 \times 158 = 31,6g\\
{m_{KCl{O_3}}} = n \times M = 0,1 \times 122,5 = 12,25g
\end{array}\)