Question

Two lab groups dissolved 6g of cupric sulfate pentahydrate in approximately 50 mL of warm water. After it dissolved, they added 1.12g of iron fillings. Student A predicted the reaction would produce 1.28g of copper, while student B predicted there would be 1.54g of copper. Which student, A or B, do you agree with? Explain your reasoning with a short essay.

Answer 1

Ans. Moles of CuSO₄.5H₂O taken = Mass / Molar mass

                                                = 6.0 g / (249.686 g/ mol)

                                                = 0.02403 mol

Since 1 mol CuSO₄.5H₂O contains 1 mol Cu-atom-

The moles of Cu-atom in 6.0 g sample = 0.02403 mol

# Moles of Fe taken = 1.12 g / (55.847 g) = 0.02005 mol

# Balanced reaction: 3 CuSO₄(aq) + 2 Fe(s) ------> Fe₂(SO₄)₃(aq) + 3 Cu(s)

Theoretical molar ration of reactants = CuSO₄ : Fe = 3 : 2

Experimental molar ratio of reactants = CuSO₄ : Fe

= 0.02403 mol : 0.02005 mol = 2.4 : 2

Comparing the theoretical and experimental molar ratios, the moles of CuSO₄ (= moles of CuSO₄.5H₂O) is lss than theoretical moles of 3 whereas that of Fe is kept constant at 2.0 moles.

Hence, CuSO₄ is the limiting reactant.

# The formation of product follows the stoichiometry of limiting reactant.

According to the stoichiometry of balanced reaction, 3 mol CuSO₄ produces 3 mol Cu.

Therefore,

            Moles of Cu produced = 0.02403 mol = moles of CuSO₄.5H₂O =

Now,

            Mass of Cu produced = 0.02403 mol x (63.546 g/ mol)

                                                = 1.53 g

# Conclusion: The result of student B is in agreement with experimental calculated values.