An iron plate was placed in the copper (II) sulfate solution. At the end of the reaction, the plate was removed

An iron plate was placed in the copper (II) sulfate solution. At the end of the reaction, the plate was removed, and a solution of barium nitrate was added dropwise to the resulting greenish solution until the formation of a precipitate ceased. The precipitate was filtered off, the solution was evaporated, and the dry salt remaining after evaporation was calcined in air. This formed a red-brown powder, which was treated with concentrated hydroiodic acid. Write the equations for the four reactions.

1) CuSO4 + Fe = + Cu (ferrous sulfate is a green solution and copper on the surface of the plate);
2) FeSO4 + Ba (NO3) 2 = BaSO4 + Fe (NO3) 2 (barium sulfate – white precipitate, solution of iron nitrate (2));
3) after filtration, a solution of iron nitrate remains (2), if you evaporate the water, we get dry salt, which decomposes when heated:
2 Fe (NO3) 2 = 2 FeO + 4 NO2 + O2 (iron oxide (2) – red-brown powder, nitrogen oxide (4), oxygen);
4) FeO + 2 HJ = FeJ2 + H2O (iron (2) iodide and water).