a, PTPƯ:
2KMnO4 \(\underrightarrow{t^o}\) K2MnO4 + MnO2 + O2
b, nO2 = \(\dfrac{2,8}{22,4}\)= 0,125 mol
mKMnO4 = 2nO2 = 0,25 mol
=> mKMnO4 cần dùng = 0,25.158 = 39,5 g
nK2MnO4 = nMnO2 = nO2 = 0,125
=> mK2MnO4 = 0,125.197 = 24,625 g
=> mMnO2 = 0,125.87 = 10,875 g