\(\begin{array}{l}
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
n{O_2} = \dfrac{{3,248}}{{22,4}} = 0,145\,mol\\
hh:KMn{O_4}(a\,mol),KCl{O_3}(b\,mol)\\
158a + 122,5b = 17,7\\
0,5a + 1,5b = 0,145\\
= > a = 0,05;b = 0,08\\
\% mKMn{O_4} = \dfrac{{0,05 \times 158}}{{17,7}} \times 100\% = 44,63\% \\
\% mKCl{O_3} = 100 - 44,63 = 55,37\% \\
n{K_2}Mn{O_4} = nMn{O_2} = 0,025\,mol\\
nKCl = nKCl{O_3} = 0,08\,mol\\
mT = 0,025 \times 197 + 0,025 \times 87 + 0,08 \times 74,5 = 13,06\% \\
\% m{K_2}Mn{O_4} = \dfrac{{0,025 \times 197}}{{13,06}} \times 100\% = 37,71\% \\
\% mMn{O_2} = \dfrac{{0,025 \times 87}}{{13,06}} \times 100\% = 16,65\% \\
\% mKCl = 100 - 37,71 - 16,65 = 45,64\%
\end{array}\)
