Fe(s) + CuSO4(aq) <===> Cu(s) + FeSO4(aq)

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 : The original concentration of copper (II) sulfate in the sample is,

Explanation :

Molar mass of Cu = 63.5 g/mol

First we have to calculate the number of moles of Cu.

Number of moles of Cu =

Now we have to calculate the number of moles of

Number of moles of Cu = Number of moles of

Number of moles of =

Now we have to calculate the molarity of

Now put all the given values in this formula, we get:

To change mol/L into g/L, we need to multiply it with molar mass of

Molar mass of = 159.609 g/mL

Concentration in g/L =

Thus, the original concentration of copper (II) sulfate in the sample is,

you gotta add the following molecules so we can give u an answer

explanation: