a) Gọi số mol KMnO4, KClO3 là a, b (mol)
=> \(\left\{{}\begin{matrix}n_K=a+b\left(mol\right)\\n_{Mn}=a\left(mol\right)\\n_{Cl}=b\left(mol\right)\\n_O=4a+3b\left(mol\right)\end{matrix}\right.\)
Có \(n_O=\dfrac{9}{14}\Sigma_n\)
=> \(4a+3b=\dfrac{9}{14}\left(a+b+a+b+4a+3b\right)\)
=> \(\dfrac{1}{7}a-\dfrac{3}{14}b=0\) (1)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
a------------------------------->0,5a
2KClO3 --to--> 2KCl + 3O2
b------------------>1,5b
=> \(0,5a+1,5b=\dfrac{10,08}{22,4}=0,45\) (2)
(1)(2) => a = 0,3 (mol); b = 0,2 (mol)
=> m = 0,3.158 + 0,2.122,5 = 71,9 (g)
b) \(\left\{{}\begin{matrix}\%m_{KMnO_4}=\dfrac{0,3.158}{71,9}.100\%=65,925\%\\\%m_{KClO_3}=\dfrac{0,2.122,5}{71,9}.100\%=34,075\%\end{matrix}\right.\)