A solution of $C u S O_{4}$ was kept in an iron pot. After few days the iron pot was found to have a number of holes in it. The balance equation of the reaction involve is:

2 F e + C u S O_{4} \to F e_{2} {(S O_{4})}_{3} + C u

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

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3 F e + C u S O_{4} \to F e_{3} (S O_{4})_{4} + C u

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F e + C u S O_{4} \to F e_{2} S O_{4} + C u

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Solution

The correct option is B $F e + C u S O_{4} \to F e S O_{4} + C u$
Iron(Fe) being more reactive than copper(Cu) displaces latter from $C u S O_{4}$ to form $F e S O_{4}$ . The reaction can be written as:
$F e + C u S O_{4} \to F e S O_{4} + C u$

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{CuSO}_{4}

C u S O_{4}

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F e S O_{4} + Z n \to Z n S O_{4} + F e

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]

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