Iron weighing 1.12 g completely reacted with a solution of copper (II) sulfate.

Iron weighing 1.12 g completely reacted with a solution of copper (II) sulfate. Find the mass of the copper precipitate formed. What amount of iron (III) sulfate was obtained?

Given:
m (Fe) = 1.12 g.
m (draft) -?
n (FeSO4) -?

Decision:
1) Fe + CuSO4 = FeSO4 + Cu
2) n (Fe) = 1.12 / 56 = 0.02 mol
3) at ur.r. n (Fe) = n (Cu) = n (FeSO4) = 0.02 mol
4) n (Cu) = 0.02 * 64 = 1.28 g.
Cu – sediment
Answer: 1.28g, 0.02 mol