10g of Fe powder was mixed with a CuSO4 solution which contained 70g of CuS

Question

10g of Fe powder was mixed with a

C u S O_{4}

solution which contained 70g of

C u S O_{4}

. 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

C u S O_{4}

. 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 +

C u S O_{4}

\to

F e S O_{4}

+ Cu
From the equation, 1 mole of Fe =1 mole of Cu

∴

55.85g

\approx

63.6g of Cu
∴ 10g of Fe will displace 11.39g of Cu if available.
The solution has 70g of

C u S O_{4}

1 mole

C u S O_{4}

has 63.6g of Cu
i.e., [63.6 + 32 + 16(4)]g of

C u S O_{4}

= 63.6g of Cu
159.6g of

C u S O_{4}

= 63.6g of Cu
70g of

C u S O_{4}

= 27.76g of Cu

∴

10 g = 10

\times

\frac{63.6}{55.85}

= 11.39 g of Cu

∴

In the

C u S O_{4}

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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