ippie

ippie
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What percentage of iron is present in the ore, if a 1.120 g sample requires 25.00 ml of 0.050 M KMnO4 to fully oxidize all of the dissolved Fe2+?

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

A. 5 * (1000 / 25.00) * 0.050 * 55.85 * (1 / 1.120) * 100 %
B. 5 * (1000 / 25.00) * 0.050 * 55.85 * 1.120 * 100 %
C. 5 * (0.050 / 1000) * 25.00 * 55.85 * 1.120 * 100 %
D. 5 * (25.00 / 1000) * 0.050 * 55.85 * (1 / 1.120) * 100 %
E. 5 * (25.00 / 1000) * 0.050 * 55.85 * 1.120 * 100 %


Answer is D.
I don't even understand what they are asking.
Please, explain me. anybody?
Thank you~~~
 

KTDMD

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ippie said:
What percentage of iron is present in the ore, if a 1.120 g sample requires 25.00 ml of 0.050 M KMnO4 to fully oxidize all of the dissolved Fe2+?

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

A. 5 * (1000 / 25.00) * 0.050 * 55.85 * (1 / 1.120) * 100 %
B. 5 * (1000 / 25.00) * 0.050 * 55.85 * 1.120 * 100 %
C. 5 * (0.050 / 1000) * 25.00 * 55.85 * 1.120 * 100 %
D. 5 * (25.00 / 1000) * 0.050 * 55.85 * (1 / 1.120) * 100 %
E. 5 * (25.00 / 1000) * 0.050 * 55.85 * 1.120 * 100 %


Answer is D.
I don't even understand what they are asking.
Please, explain me. anybody?
Thank you~~~
Let me try putting in the units and see if this will help:

{[(5 mol Fe2+/1 mol MnO4-) * 25 ml reagent * (0.050 mol MnO4-/1000 ml reagent) * (55.85 g Fe2+/1 mol Fe2+)] over 1.120 g ore sample} * 100%

(Note: Maximize your browser screen so the above equation stays in one line and cancel out those units before associating those terms such that they will conform to D.)
 

asckwan

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KTDMD gave a great solution, but just in case you are like me, who needs to see things really in pieces, this is kind of how I approached the problem. I broke it complete down into parts then put them together.

The question is asking for a percent of a whole, so the whole thing weighs 1.12g, and you want to find grams of Fe. So keep in mind you are looking for Fe2+ g first.

Part 1: So we have .05M of KMnO4 and 25mL of it. So .05M KMnO4 x .025L = # of moles of KMnO4.

Part 2: We have 5 moles of Fe2+ for every 1 mole of KMnO4, so in order to get the correct amount of Fe2+ we need to multiply the number of moles of KMnO4 by the number of moles of Fe2+. Therefore, we need to multiply by 5.

Part 3: We need GRAMS of Fe2+, so we should multiply the # of moles of Fe2+ by it's weight, 55.85.

Part 4: To make it into percent of whole you will take Grams of Fe2+ and divide by Grams of the whole thing = 1.12

Part 5: put it all together

(Moles of KMnO4) x (5 moles Fe/1 mole KMnO4) x (Fe2+ grams) / (Sample weight) x 100

(.05M x .025L) x (5/1) x (55.85g) / 1.12 x 100

Hope this helps!!