To attack a complex problem, try a 'divide and conquer' strategy as follows:
Step 1: Choose your target. You want to get a percentage of iron in the ore. If you had the mass of iron in the ore and the mass of the ore, you'd have your answer. The mass of ore is given, but the mass of iron isn't: your target is g Fe.
Step 2: Examine the given information. The problem gives you several pieces of information; look at each one closely and try to decide how it is related to your target.
0.206 g KMnO_{4} in 100.0 mL KMnO_{4} solution | This can be used to relate moles of KMnO_{4} to L KMnO_{4} solution. |
0.238 g iron ore in 50.0 mL distilled water | The amount of ore will be necessary to compute the percentage of the iron in the ore. (By the way, water won't work as a solvent here. I believe this part of the problem should specify that all of the iron in the ore is converted to Fe^{2+}(aq)). |
1.7 mL of H_{2}SO_{4} | This tells you the redox reaction between the KMnO_{4} and the Fe^{2+} is occuring in an acidic solution. The amount of H_{2}SO_{4} is not crucial here, since the limiting reagent in the titration will be the Fe^{2+}. |
10.3 mL of KMnO_{4} solution at endpoint | This tells you how much KMnO_{4} is required to completely react with all of the Fe^{2+} from the ore sample. (KMnO_{4} is deep purple and all the other species in the titration are nearly colorless. Just one drop of excess permanganate imparts a definite pink tinge to the solution.) |
Step 3: Relate the target to information given in the problem. The key piece in the remainder of the problem is relating the amount of KMnO_{4} added to the amount of iron in the ore:
Once you have the balanced equation, you can look at the coefficients and convert mol KMnO_{4} to mol Fe. This gives you a 'bridge' from your starting point to the target:
The mole-to-mole relationship cuts the problem up into 3 easier pieces:
Step 4: Perform and check the calculation.Verify that all the units in the calculation we set up in the previous step 'cancel' consistently. You should get an iron mass that is some fraction of 0.238 g (the iron can't weigh more than the ore it came from, right?) Use the grams of iron and the grams of ore to obtain the percentage of iron in the ore.
Copyright © 1997-2010 by Fred Senese
Comments & questions to fsenese@frostburg.edu
Last Revised 08/17/15.URL: http://antoine.frostburg.edu/chem/senese/101/moles/faq/print-titration-fe-kmno4.shtml