\(\text{PTHH}:\,\,\, 2KMnO_4\xrightarrow{t^o} K_2MnO_4+MnO_2+O_2\\n_{KMnO_4}=\dfrac{m_{KMnO_4}}{M_{KMnO_4}}=\dfrac{47,4}{158}=0,3(mol)\\\text{Theo pt ta có tỉ lệ}:\,\, 2:1:1:1\\→n_{O_2}=0,15(mol)\\→m_{O_2}=n_{O_2}.M_{O_2}=0,15.32=4,8(g)\\\text{Vậy thu được}\,\, 4,8g \,\,Oxi\,\, \text{khi nhiệt phân}\,\, 47,4\,\, KMnO_4\)