2KCLO3->2KCl+3O2
245g 149g
a g \(\dfrac{149.a}{245}\)g
2KMnO4=>K2MnO4+MnO2+O2
316g 197g 87g
b g \(\dfrac{197b}{316}\)g \(\dfrac{87b}{316}g\)
=>\(\dfrac{a}{b}=\dfrac{\dfrac{149a}{245}}{\dfrac{197b}{316}+\dfrac{87b}{316}}=\dfrac{\dfrac{149a}{245}}{\dfrac{284b}{316}}=\dfrac{149a.316}{245.284b}=\dfrac{0,68a}{b}=>\dfrac{a}{b}=0,68\)