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,

answer is a

explanation:

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.