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.

The original concentration of copper sulfate is 0.56 g/L

Explanation:

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

    .....(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:

The given chemical equation follows:

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 = 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:

Mass of copper sulfate = 0.223 g

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

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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :