\(a.2KMnO_4-^{t^o}\rightarrow K_2MnO_4+MnO_2+O_2\\ b.n_{O_2}=0,3\left(mol\right)\\ n_{KMnO_4}=2n_{O_2}=0,6\left(mol\right)\\ \Rightarrow m_{KMnO_4\left(pứ\right)}=158.0,6=94,8\left(g\right)\\ TrongA:\left\{{}\begin{matrix}K_2MnO_4:0,3\left(mol\right)\\MnO_2:0,3\left(mol\right)\end{matrix}\right.vàcóthểcóKMnO_4dư\\ m_{KMnO_4\left(dư\right)}=100-\left(0,3.197+0,3.87\right)=14,8\left(g\right)\\ \Rightarrow m_{KMnO_4\left(bđ\right)}=94,8+14,8=109,6\left(g\right)\\ c.\%m_{KMnO_4\left(dư\right)}=\dfrac{14,8}{100}.100=14,8\%\\ \%m_{K_2MnO_4}=\dfrac{0,3.197}{100}.100=59,1\%\\ \%m_{MnO_2}=26,1\%\)
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