\(\begin{array}{l}
1)\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
C + {O_2} \to C{O_2}\\
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
2){\rm{ox}}it{\rm{ax}}it:S{O_2},{N_2}{O_3},C{O_2}\\
{\rm{ox}}it\,bazo:CaO,F{e_2}{O_3},CuO\\
3)\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
n{O_2} = \frac{{48}}{{32}} = 1,5\,mol\\
= > nKCl{O_3} = 1\,mol\\
mKCl{O_3} = 1 \times 122,5 = 122,5g
\end{array}\)