Your family is in the business of processing iron from iron ore and after taking

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 KMnO4 to reach end point.

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

The selling price of the raw iron as produced above is $0.80/lb and there is an additional $400.00 cost associated with entire process for the truck load of ore. How much should your family business pay to make 40% profit from this business? Assume none of the byproducts have any commercial value.

Explanation / Answer

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

Moles of KMnO4 required per gram of ore = molarity*volume of solution in litres = 0.086*0.025 = 2.15*10-3

Thus, moles of Fe2+ present in 1 g of the ore = 5*moless of KMnO4 required = 0.01075

Therefore total moles of Fe present in the ore = 4000*450*0.01075 = 19350 (1 lb = 0.45 kg)

Thus, total mass of Fe present = moles*molar mass of Fe = 19350*56 = 1083600 g

% mass recovered = 0.8*1083600 = 866880 g

Total selling price = 0.8*866880/450 = $ 1541.12

Now, C.P = S.P - 0.4*C.P

or, 1.4*C.P = S.P

or, C.P = $ 1100.8

Thus, to make 40% profit the family should pay 1100.8 - 400 = $ 700.8 to the customer

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