1)If 36 mL of KMnO4 solution are required to oxidize 25 mL of 0.02932 M NaC204 solution, what is the concentration of the solution?

2)A Certain brand of iron supplement contains FeSO4-7H2O with miscellaneous binders and fillers. Suppose 22.93 mL of the KMn04 solution used in question one above are needed to oxidize Fe2+ to Fe3+ in a 0.4927 g pill. What is the mass % of FeSO4-7H2O(MM 278.03 g mol) in the pill?

I understand part 1, I think..

I have 0.02932 M x 0.025 L = 0.000733 mol Na2C2O4

Then,
0.000733 mol x (2/5) = 0.0002932/0.036 L
=0.008144 M KMnO4

But for number 2, I have no idea. Someone help plz!

3 answers

1 is ok. I'm surprised you can do that with no trouble but have a problem with #2.
The problem state Fe^+2 is oxidized to Fe^+3; therefore you know the ratio is 5Fe to 1 MnO4.
mL x M MnO4 = mmoles MnO4.
Convert mmole MnO4 to mmoles Fe.
Convert mmols Fe to mmoles FeSO4.7H2O and convert that to grams.
g = moles x molar mass (or mmoles x mmolar mass)
Then %FeSO4.7H2O = (mass FeSO4.7H2O/mass sample)*100 = ??

Thanks

melting point of mno4