Suppose 22.93mL of KMnO4 solution used in the question above (KMnO4 solution in above question had concentration of 0.02036M)are needed to oxidize Fe(2+) to Fe(3+) in a 0.4927g pill. What is the mass percent of FeSO4*7H2O (molar mass 278.03g/mol) in the pill?

I know I am supposed to use the KMnO4 to find mols of Fe(II). But after doing that I am unsure as to how i should use Fe(II) to find the mass of FeSO4*7H2O

moles Fe+2 is the same as the moles of FeSO4.7H2O. mass of that is molmass * numbermoles

To find the mass percent of FeSO4·7H2O in the pill, you first need to determine the number of moles of Fe(II) that reacted with the KMnO4 solution.

1. Calculate the number of moles of KMnO4 used:
Given volume of KMnO4 solution = 22.93 mL
Molarity of KMnO4 solution = 0.02036 M

Moles of KMnO4 = Volume (in L) x Concentration (in mol/L)
= 0.02293 L x 0.02036 mol/L

2. The balanced chemical equation for the reaction between KMnO4 and Fe(II) is:
5Fe(II) + 4KMnO4 + 8H2SO4 -> 5Fe(III) + 4MnSO4 + K2SO4 + 8H2O

According to the stoichiometry, the ratio of Fe(II) to KMnO4 is 5:4. Therefore, the moles of Fe(II) can be calculated as:
Moles of Fe(II) = (5/4) x Moles of KMnO4

3. Determine the mass of FeSO4·7H2O:
Given mass of the pill = 0.4927 g
Molar mass of FeSO4·7H2O = 278.03 g/mol

Moles of FeSO4·7H2O = Mass (in g) / Molar mass (in g/mol)

4. Finally, calculate the mass percent of FeSO4·7H2O in the pill:
Mass % = (Moles of FeSO4·7H2O / Total moles of all components in the pill) x 100

Total moles of all components in the pill = Moles of Fe(II) + Moles of FeSO4·7H2O

By following these steps, you should be able to find the mass percent of FeSO4·7H2O in the pill.