Question

Fe(s)+CuSO4(aq)========Cu(s)+FeSO4(aq)*Note both 4's are subscripts and the equal signs represent an arrow.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 digits. Answer in g/L.

Answer 1

Answer: The original concentration of copper sulfate is 0.56 g/L

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

Given mass of copper = 89 mg = 0.089 g   (Conversion factor: 1 g = 1000 mg)

Molar mass of copper = 63.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of copper}=\frac{0.089g}{63.5g/mol}=0.0014mol$

The given chemical equation follows:

$Fe(s)+CuSO_4(aq.)\rightarrow Cu(s)+FeSO_4(aq.)$

By Stoichiometry of the reaction:

1 mole of copper metal is produced by 1 mole of copper sulfate

So, 0.0014 moles of copper metal will be produced by = $\frac{1}{1}\times 0.0014=0.0014mol$ of copper sulfate

Now, calculating the mass of copper sulfate from equation 1, we get:

Molar mass of copper sulfate = 159.6 g/mol

Moles of copper sulfate = 0.0014 moles

Putting values in equation 1, we get:

$0.0014mol=\frac{\text{Mass of copper sulfate}}{159.6g/mol}\\\\\text{Mass of copper sulfate}=(0.0014mol\times 159.6g/mol)=0.223g$

• Calculating the original concentration of copper sulfate:

Mass of copper sulfate = 0.223 g

Volume of copper sulfate = 400 mL = 0.400 L    (Conversion factor: 1 L = 1000 mL)

$\text{Original concentration of copper sulfate}=\frac{0.223g}{0.400L}=0.56g/L$

Hence, the original concentration of copper sulfate is 0.56 g/L