PTHH: \(2KMnO_4-t^o->K_2MnO_4+MnO_2+O_2\)
\(m_{KMnO_4\left(pư\right)}=30,8.90\%=27,72\left(g\right)\)
=> \(n_{KMnO_4}=\dfrac{27,72}{158}=0,175\left(mol\right)\)
Theo PT ta có: \(n_{O_2}=\dfrac{1}{2}.n_{KMnO_4}=\dfrac{1}{2}.0,175=0,0875\left(mol\right)\)
=> \(V_{O_2}=0,0875.22,4=1,96\left(l\right)\)