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!

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

To find the mass % of FeSO4-7H2O in the pill, you need to calculate the number of moles of FeSO4-7H2O and then convert it to mass. Here's how you can approach the problem:

1) Determine the number of moles of KMnO4 used:
From the first question, we know that it took 36 mL of KMnO4 solution to oxidize 25 mL of Na2C2O4 solution. You have already calculated the concentration of the KMnO4 solution to be 0.008144 M. Use the formula C x V = n to calculate the number of moles of KMnO4 used:
0.008144 M x 0.036 L = 0.0002932 mol KMnO4

2) Using the balanced chemical equation, determine the molar ratio between KMnO4 and FeSO4-7H2O:
The balanced equation for the reaction between KMnO4 and FeSO4-7H2O is:
10 FeSO4-7H2O + 2 KMnO4 + 8 H2SO4 -> 10 Fe2(SO4)3 + 2 MnSO4 + 8 H2O + 5 K2SO4

From the equation, we see that 2 moles of KMnO4 react with 10 moles of FeSO4-7H2O. Therefore, the mole ratio is 2:10 or 1:5.

3) Calculate the number of moles of FeSO4-7H2O:
Since the mole ratio between KMnO4 and FeSO4-7H2O is 1:5, we can calculate the number of moles of FeSO4-7H2O as follows:
0.0002932 mol KMnO4 x (5 mol FeSO4-7H2O / 2 mol KMnO4) = 0.000733 mole FeSO4-7H2O

4) Calculate the mass of FeSO4-7H2O:
Now we can convert the number of moles of FeSO4-7H2O to mass using the molar mass (MM) of FeSO4-7H2O, which is given as 278.03 g/mol:
0.000733 mol FeSO4-7H2O x 278.03 g/mol = 0.203 g FeSO4-7H2O

5) Calculate the mass % of FeSO4-7H2O in the pill:
Finally, we can calculate the mass % of FeSO4-7H2O by dividing the mass of FeSO4-7H2O by the mass of the pill and multiplying by 100:
(0.203 g FeSO4-7H2O / 0.4927 g pill) x 100 = 41.24%

Therefore, the mass % of FeSO4-7H2O in the pill is approximately 41.24%.

melting point of mno4