Assignment should be typed double spacing Balanced equation must include all sta

Question

Assignment should be typed double spacing Balanced equation must include all state symbols (there is no partial credit for an incomplete Your family is in the business of processing iron from iron ore and after taking CHEM 1411, 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 Fc2' before redox titration. The oxidizing agent used is a standardized 0.086M potassium per magnetic (KMn04). A solution of 1.00 g iron ore required 25.00 mL of 0.086M KMnO, to reach end point. The yield of iron from the iron ore in the entire process is 80%. Write a balanced net ionic equation for the redox titration with KMn04. Calculate the mass of iron in 1.00 g iron ore What is the percent of iron in the ore? Calculate the theoretical yield of iron from the truck load. What is the actual yield of iron from the truck load? Bonus Question The selling price of the raw iron as produced above is S0.80 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

1) After reduction of iron ore to Fe2+, the overall iron in the ore will be in the form of F2+ only.

the balanced equation for the reaction with MnO4- is as below

5 Fe2+(aq) + MnO4-(aq) + 8H+(aq) ---> 5Fe3+(aq) + Mn2+(aq) +4H2O(l)

2) from the above equation we can say that 1 mole of KMnO4 will require 5 moles of iron

    from given data 1 g of iron ore requires 25ml of 0.086M KMnO4

    No. of moles of KMnO4 = 0.086 mole/Litre x 0.025 L = 0.00215 moles

    Hence No. of moles of iron = 5x 0.00215 =0.01075 moles of iron present in 1 g of iron ore

    Mass of iron = 0.01075 moles x 55.8 g/mole = 0.59985 g

3) percent of iron = (weight of iron / weight of ore) x 100 = (0.59985g / 1g) x 100 = 59.985 %

4) Truck load received = 4000lb = 4000 x 453.592 g = 1814368 g (since 1 lb = 453.592g)

    If we applied the above calculated percen we will get theoritical yield

    = 1814368 g x 59.985 / 100 = 1088348 g = 2399.4 lb

5) From given data, the % yield = 80% throught the process

    Then the actual yield for 4000lb = 4000 x 80 /100 = 3200 lb = 1451494.4 g

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.