2KMnO4-->K2MnO4+MnO2+O2
2O2+3Fe---.>Fe3O4
n\(_{KMnO4}=\)\(\frac{94,8}{158}=0,6\left(mol\right)\)
Theo pthh
n\(_{O2}=\frac{1}{2}n_{KMnO4}=0,3\left(mol\right)\)
n\(_{Fe}=\frac{22,4}{56}=0,4\left(mol\right)\)
=> O2 dư
Theo pthh
n\(_{Fe3O4}=\frac{1}{3}n_{Fe}=0,133\left(mol\right)\)
m\(_{Fe304}=0,133.232=30,86\left(g\right)\)
Ta có : nKMnO4 = 0,6(mol)
PTHH: 2KMnO4--->K2MnO4+MnO2+O2
3Fe+2O2--->Fe3O4
=>nO2 = 0,3(mol)
=>nFe = 0,4(mol)
Lập tỉ lệ lên ta thấy O2 dư
=>nFe3O4 = 0,133 (mol)
=>mFe3O4 = 30,86g