\(n_{P_2O_5} = \dfrac{28,4}{142} = 0,2(mol)\)
4P + 5O2 \(\xrightarrow{t^o}\) 2P2O5
0,4...........0,5...............0,2.....................(mol)
Suy ra:
\(m_P = 0,4.31 = 12,4(gam)\\ m_{O_2} = 0,5.32 = 16(gam)\)
\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4\ pư} = 2n_{O_2} = 1(mol)\\ \Rightarrow m_{KMnO_4\ đã\ dùng} = \dfrac{1.158}{90\%} = 175,5(gam)\)