Two lab groups dissolved 6g of
Two lab groups dissolved 6g of cupric sulfate pentahydrate inapproximately 50 mL of warm water. After it dissolved, they added1.12g of iron fillings. Student A predicted the reaction wouldproduce 1.28g of copper, while student B predicted there would be1.54g of copper. Which student, A or B, do you agree with? Explainyour reasoning with a short essay.
Answer:
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 molCu-atom-
The moles of Cu-atom in 6.0 g sample = 0.02403mol
# Moles of Fe taken = 1.12 g / (55.847 g) = 0.02005mol
# Balanced reaction: 3 CuSO4(aq) + 2Fe(s) ——> Fe2(SO4)3(aq) + 3Cu(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, themoles of CuSO4 (= moles ofCuSO4.5H2O) is lss than theoretical moles of3 whereas that of Fe is kept constant at 2.0 moles.
Hence, CuSO4 is the limitingreactant.
# The formation of product follows the stoichiometry of limitingreactant.
According to the stoichiometry of balanced reaction, 3 molCuSO4 produces 3 mol Cu.
Therefore,
Moles of Cu produced = 0.02403mol = 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 Bis in agreement with experimental calculated values.
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