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]

10.39

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11.39

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12.39

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Solution

The correct option is B
11.39

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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Similar questions

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]

Q. Chemical equation for the reaction between

C u S O_{4}

solution and an iron nail is given below.

C u S O_{4} + F e \to F e S O_{4} + C u

(a) Write the reduction reaction taking place here.
(b) Give reason for the displacement of Cu by Fe from

C u S O_{4}

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$E_{F e^{2 +} / F e}^{\circ} = - 0.44 V, E_{C u^{2 +} / C u}^{\circ} = 0.34 V, E_{A g^{2 +} / g}^{\circ} = 0.80 V$ . Will Fe displace Cu from $C u S O_{4}$ solution? Will Ag displace Cu from $C u S O_{4}$ solution?

1. Yes, Yes

2. No, No

3. Yes, No

4. No, Yes

Q. The reaction

Z n + C u S O_{4} = C u + Z n S O_{4}

goes completely to the right. In one experiment

10 g

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C u O

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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]

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]