\(\begin{array}{l}
4P + 5{O_2} \to 2{P_2}{O_5}\\
4Na + {O_2} \to 2N{a_2}O\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}\\
12)\\
a)2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
b)\\
nKMn{O_4} = \frac{{15,8}}{{158}} = 0,1\,mol\\
= > n{O_2} = 0,05\,mol\\
V{O_2} = 0,05 \times 22,4 = 1,12l\\
c)\\
nS = \frac{{3,2}}{{32}} = 0,1\,mol\\
S + {O_2} \to S{O_2}\\
nS{O_2} = n{O_2} = 0,05\,mol\\
mS{O_2} = 0,05 \times 64 = 3,2g
\end{array}\)