\(V_{O_2\left(thu.được\right)}=28=0,1=2,8\left(l\right)\)
=> \(V_{O_2\left(sinh.ra\right)}=\dfrac{2,8.100}{80}=3,5\left(l\right)\)
=> \(n_{O_2\left(sinh.ra\right)}=\dfrac{3,5}{22,4}=0,15625\left(mol\right)\)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,3125<------------------------0,15625
=> mKMnO4 = 0,3125.158 = 49,375 (g)