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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]


A

10.39

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B

11.39

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C

12.39

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D

13.39

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Solution

The correct option is B

11.39


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.


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