Question

The Fe2 (55.845 g/mol) content of a 2.271 g steel sample dissolved in 50.00 mL was determined by tiration with a standardized 0.100 M potassium permanganate (KMnO4, 158.034 g/mol) solution. The titration required 28.05 mL to reach the end point. What is the concentration of iron in the steel sample? Express your answer as grams of Fe per grams of steel (g Fe2 / g steel).

Answer 1

I don't like the Fe2 business. Probably you mean Fe^2+ and even though that is a perfect way of writing it the Fe2 bit makes one think iron is diatomic and it isn't.
5Fe^2+ + KMnO4 ==> 6Fe^3+ + Mn^2+
(I know the equation is balanced but the redox part is and we aren't worried about the spectator ions/molecules (H^+ and H2O). You can put those in if you wish.
mols KMnO4 = M x L = ?
mols Fe^2+ = 5 x that(5 times that)
g Fe = mols Fe x atomic mass Fe = ? = grams Fe in the 2.271 g sample of steel.
So convert that from g Fe in 2.271g steel to grams Fe in 1 g steel.

Answer 2

So I'm I correct in doing
mols KMnO4 = 0.1M x 0.05L = 0.005
mols Fe^2+ = 5 x 0.005 = 0.025
g Fe = 0.025mol x 55.845g/mol = 1.396125 = grams Fe in the 2.271 g sample of steel.
So convert that from g Fe in 2.271g steel to grams Fe in 1 g steel.
So to do this last part do I take the 1.396125/2.271g? To find what is in the 1g sample?
because I got 0.6147g and that is incorrect.

Answer 3

mols KMnO4 is M x L = 0.02805 x 0.1 =? You dissolved the sample in 50 mL, the titration with KMnO4 was 28.05 mL. Correct that step and follow through with the rest of the problem The set up looks ok for the rest of it. I ran through it and obtained approx 0.35 g Fe/g sample.