gchem question

[tooth123]

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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





Not sure how to solve this problem, help would be appreciated... thanks!

 

[UndergradGuy7]

Stoichiometry. Just find out the grams of Fe2+ that reacts with MnO4- and then find the percent.

You have 25 mL of KMnO4 that is 0.050 M. So you have 0.025 L x 0.050 M = mols of MnO4-. From the reaction equation they give you you know that MnO4- reacts with Fe2+ in a ratio 1MnO4-:5Fe2+. So mols of MnO4- x (5 Fe2+/1 MnO4-). Now you have mols of Fe2+. Convert to grams so (x mols Fe2+) x ( molar mass of Fe/ 1 mol) = mass of Fe2+. Then percent is (mass/1.120) * 100.

Or the whole equation (0.050)*(25/1000)*(5/1)*(molar mass Fe/1mol)*(1/1.120)*100 = percent.