. Your family business just received a 4000 lb truck load of iron ore from a new
ID: 928673 • Letter: #
Question
.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%
.A) Write a balanced net ionic equation for the redox titration with KMnO4
B) Calculate the mass of iron in 1.00 g iron ore
C) What is the percent of iron in the ore?
D) Calculate the theorectical yield of iron from the truck load.
E) What is the actual yield of iron from the truck load?
Explanation / Answer
Convert units 4000 lb x 0.45359 lb/kg = 1814.4 kg
a.
MnO4(aq)+5Fe2+(aq)+8H+(aq) = 5Fe3+(aq)+Mn2+(aq)+4H2O(l)
b.
mFe = 0.025L x 0.086mol/L x 5 x 55.845 gFe/mol =
= 0.600g
c.
100 x 0.600 g/1.00g = 60.0%
d.
1814.4 kg x 60/100 = 1088 kg Fe may be obtained (without losses during the process)
e.
80/100 x 1088 kg = 871 kg may be obtained (with 20% losses)
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