2KMnO4 → K2MnO4 + MnO2 + O2по уравнению реакции:
n(KMnO4) / 2 = n(MnO2) = n(O2) = n(K2MnO4);
n(KMnO4) = m(KMnO4) / M(KMnO4) = 63.2 / (39+55+16*4) = 63.2 / 158 = 0.4 (mol);
n(MnO2) = n(O2) = n(K2MnO4) = n(KmO4) / 2 = 0.4 / 2 = 0.2 (mol);
V(O2)=n(O2)*Vm = 0.2*22.4 = 4.48 (л);
m(MnO2) = n(MnO2)*M(MnO2) = 0.2*(55+16*2) = 0.2*87 = 17.4 (g);
m(K2MnO4)=n(K2MnO4)*M(K2MnO4) = 0.2*(39*2+55+16*4) = 0.2*197 = 39.4 (g)
w = V(O2)(prakt)*100% / V(O2)(teor) = 3.36*100% / 4.48 = 75%
Answer: w = 75%....m(MnO2) = 17.4 g.....m(K2MnO4)=39.4 g