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ChemistryMole ConceptLaws of Chemical CalculationsMedium2 minQB

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A 4.47 g sample of a mixture of $CuCl_{2}$ and $CuBr_{2}$ was dissolved in water and mixed thoroughly with a 5.74 g portion of AgCl. After the reaction the solid, a mixture of AgCl and AgBr, was filtered, washed, and dried. Its mass was found to be $6.63 g (Cu = 63.5, Ag = 108, Br = 80)$

The mass percent of $CuBr_{2}$ in original mixture is

Options:

Answer:

Solution:

Let CuBr₂ = $x$ g and CuCl₂ = $(4.47 - x)$ g.

CuBr₂ reacts with AgCl: CuBr₂ + 2AgCl → CuCl₂ + 2AgBr.

In this reaction, each mole of CuBr₂ replaces AgCl with AgBr. The mass increase is $(188 - 143.5) = 44.5$ g per mole.

The increase in precipitate mass is $6.63 - 5.74 = 0.89$ g.

Moles of CuBr₂ = $\frac{0.89}{44.5} = 0.02$ mol.

The mass of CuBr₂ = $0.02 \times 223.5 = 4.47$ g.

Therefore, the percentage of CuBr₂ = $(\frac{4.47}{8.94}) \times 100 = 50%$ .

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Laws of Chemical Calculations

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