7.80 g of copper (II) sulfate pentahydrate (CuSO4*5H2O) were dissolved in a beaker of 50 mL of water and heated slowly. 2.00 g of iron filings were added to the solution, and then the mixture was allowed to cool. The supernate was decanted, and the contents of the beaker were dried. The solid at the bottom was reddish-brown, and its mass was 2.36 g.
What was this solid? Was it iron (II) sulfate, iron (III) sulfate, or copper? Was it hydrated?
If iron (II) sulfate formed...CuSO4*5H2O(aq) + Fe(s) = Cu(s) + FeSO4(aq) + 5 H2O(g)
If iron (III) sulfate formed...3 CuSO4*5H2O(aq) + 2 Fe(s)= 3 Cu(s) + Fe2(SO4)3(aq) + 15 H2O(g)
The Attempt at a Solution
The solid was reddish-brown, so it is probably not an iron (II) compound (which are greenish). I don't believe that the solid was hydrated; water vapor rose from the beaker when it was heated (although that may just be due to the aqueous solution).
I attempted to find the theoretical yield by first finding the limiting reactant (CuSO4*5H2O) and then determining the expected mass of compounds. If the solid was copper and was anhydrous, the mass should be 1.96 g; if it was iron (III) sulfate, the mass should be 8.31 g, and if it was iron (II) sulfate it should be 4.74 g.
I'm unsure of the solid's identity. Please help.