A weighed portion of 1.12 g of iron is completely “dissolved” in a solution of copper (II) sulfate. Calculate

A weighed portion of 1.12 g of iron is completely “dissolved” in a solution of copper (II) sulfate. Calculate the mass of the copper precipitate formed. How much of the substance of iron (II) sulfate was obtained in this case?

Given:
m (Fe) = 1.12 g.

Solution:
1) Fe + CuSO4 = FeSO4 + Cu
2) n (Fe) = 1.12 / 56 = 0.02 mol
3) according to the equation n (Fe) = n (Cu) = 0.02 mol
4) m (Cu) = 0.02 * 64 = 1.28 g.
5) n (FeSO4) = n (Fe) = 0.02 mol

Answer: 1.28g, 0.02 mol