CHEMISTRY
posted by yipez .
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 FeSO47H2O 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 FeSO47H2O(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!

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