1. First, balance the Fe/MnO4 equation but correct it first. The part that matters is 5Fe^2+ + MnO4^- ==> 5Fe^3+ + Mn^2+.
a. Millimols MnO4 used = M x mL = 12.08 x 0.200 = approx 2.4.
Using the coefficients in the balanced equation, convert mmols MnO4 to mmols Fe.
That's approx 2.4 x (5 Fe/1 MnO4) = approx 12 mmols Fe = 0.012 mols.
Convert to g Fe. Approx 0.012 x 55.85 = approx 0.7 g Fe in the 10.0 mL aliquot titrated or approx 7 g Fe in the origininal sample.
Note that this gives you a larger amount of Fe than the initial sample and that isn't possible. My best guess is that the MnO4 solution is 0.02 M and not 0.2 M. Look over the numbers and make sure they are right.
b. %purity = (mass Fe/mass sample)*100 ?
An iron nail with a mass of 0.750 g was dissolved in dilute sulfuric acid and diluted to 100.0 mL volume. A 10.0 mL aliquot of this solution was then titrated with standard 0.200 M potassium permanganate solution, according to the following equation: MnO4- + H+ +Fe2+ --> Mn2+ + Fe3+ +H2O
It took 12.08 mL of the permanganate solution to oxidize all the iron.
a. Determine the mass of iron present in the original sample.
b. Given the result from (a), what was the purity of the iron nail?
c. If the nail contained a non-iron impurity that did not react with the sulfuric acid, but was oxidized by the permanganate, would the calculated purity of the nail be overestimated or underestimated?
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