Question

Your family is in the business of processing iron from iron ore and after taking Chemistry, you want to help the family business with this knowledge.

Your family business just received a 4000 lb truck load of iron ore from a new customer. You were provided with following background information:

-The iron in the ore has been reduced to Fe²⁺ before redox titration.

-The oxidizing agent used is a standardized 0.086M potassium per manganate (KMnO₄).

-A solution of 1.00 g iron ore required 25.00 mL of 0.086M KMnO₄to reach end point.

-The yield of iron from the iron ore in the entire process is 80%.

What is the Theoretical and Actual Yield from the truck load?

Answer 1

Answer – We are given, mass of iron ore = 4000 lb ,

[KMnO₄] =0.086M, volume = 25.00 mL , for 1.0 g of titration need 25.0 mL of 0.086 M KMnO₄.

yield of iron from the iron ore in the entire process = 80%

The redox reaction between Fe²⁺ and KMnO₄ is as follow

MnO₄^- + 5Fe²⁺ + 8H⁺ ----> Mn²⁺ + 5Fe³⁺ + 4H₂O

So we need to calculate moles of MnO₄^-, means moles of KMnO₄.

Moles of KMnO₄ = 0.086 M * 0.025 L

                             = 0.00215 moles

From the balanced redox reaction

1 moles of MnO₄^- = 5 moles of Fe²⁺

So, 0.00215 moles of MnO₄^- = ?

= 0.01075 moles of Fe²⁺

So, mass of Fe = 0.01075 moles * 55.845 g/mol

                         = 0.600 of Fe

So, 1.0 g of iron ore = 0.600 of Fe

Now we need to convert the mass of iron ore lb to g

1 lb = 453.59 g

So, 4000 lb= ?

= 1814360 g

So, 1.0 g of iron ore = 0.600 of Fe

Then 1814360 g of iron ore = ?

= 1089222 g of Fe

So theoretical yield of Fe = 1089222 g of Fe

                                         =2401.32 lb of Fe

We are given the percent yield = 80 % and we know formula

Percent yield = actual yield / theoretical yield * 100 %

So, actual yield = percent yield * theoretical yield / 100 %

                         = 80 % * 1089222 g of Fe / 100 %

                          = 871377.2 g of Fe

                         = 1921.06 lb of Fe