Question -10g of Fe powder was mixed with a CuSO4 solution which contained 70g of CuSO4. What is the weight of the copper obtained? [Fe = 55.85g Cu = 63.6g]
Options:
A .  10.39
B .  11.39
C .  12.39
D .  13.39
Answer: Option B : B The question talks about reaction between Fe and CuSO4. Observe that weights of the reactants are given and the weight of a product is asked. Thus, it is a weight-weight stoichiometry problem. In all such stoichiometry problems, start by balancing the reaction Fe + CuSO4→FeSO4 + Cu From the equation, 1 mole of Fe =1 mole of Cu ∴ 55.85g ≈63.6g of Cu ∴ 10g of Fe will displace 11.39g of Cu if available. The solution has 70g of CuSO4 1 mole CuSO4 has 63.6g of Cu i.e., [63.6 + 32 + 16(4)]g of CuSO4 = 63.6g of Cu 159.6g of CuSO4= 63.6g of Cu 70g of CuSO4= 27.76g of Cu ∴10 g = 10 ×63.655.85 = 11.39 g of Cu ∴ In the CuSO4 solution, 27.76g of Cu is present of which 11.39g of Cu will be precipitated out by the addition of 10g of Fe.
Submit Your Solution hear: