Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of . Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to significant digits.

Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.

Solution

Noting that the precipitate is Copper as it is the only solid by-product of this reaction.

89 mg of Copper is produced from this reaction.

We convert this into number of moles for further stoichiometric calculations

Mass of Copper = 89 mg = 0.089 g

Molar mass of Copper = 63.546 amu

Number of moles of Copper produced from the reaction = (0.089/63.546) = 0.0014005602 = 0.001401 mole

From the stoichiometric balance of the reaction,

1 mole of Copper is produced from 1 mole of Copper (II) Sulfate

0.001401 mole of Copper will be produced similarly from 0.001401 mole of Copper (II) Sulfate.

Number of moles of Copper (II) Sulfate in the original sample = 0.001401 mole

Concentration of Copper (II) Sulfate in the original sample in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = 0.001401 mole

Volume in L = (400/1000) = 0.4 L

Concentration of Copper (II) Sulfate in the original sample in mol/L = (0.001401/0.4) = 0.0035025 mol/L = 0.0035 mol/L to 2 s.f.

Concentration in g/L = (Concentration in mol/L) × (Molar Mass)

Concentration in mol/L = 0.0035025 M

Molar mass of Copper (II) Sulfate = 159.609 g/mol

Concentration of Copper (II) Sulfate in the original sample in g/L = 0.0035025 × 159.609 = 0.559 g/L = 0.56 g/L to 2 s.f

Hope this Helps!!!!

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-11-18T17:39:48+01:00

The concentration of the original copper solution is 0.035 M.

The equation of the reaction is;

Fe(s) + CuSO4(aq) -------> FeSO4(aq) + Cu(s)

Number of moles of copper obtained = 89 × 10^-3g/63.5 = 0.0014 moles

Since the reaction is 1:1, the number of moles of copper sulfate that reacted is c.

From the question, we are told that the volume of solution is 400.mL or 0.04L.

Hence, the concentration of the solution is; number of moles /volume

= 0.0014 moles/0.04L = 0.035 M

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Missing parts;

Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.