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!
see above.
To find the mass % of FeSO4-7H2O in the iron supplement pill, you can follow these steps:
1) Determine the number of moles of KMnO4 used in the reaction:
The molarity of the KMnO4 solution is given as 0.008144 M.
Using the volume, 22.93 mL (which can also be converted to 0.02293 L), and the molarity, we can calculate the number of moles of KMnO4 used:
Moles of KMnO4 = Molarity x Volume
= 0.008144 mol/L x 0.02293 L
= 0.000187 mol
2) Determine the number of moles of Fe2+ oxidized:
From the balanced chemical equation for the reaction, we know that the ratio of moles of KMnO4 to moles of Fe2+ is 5:2.
Moles of Fe2+ = (5/2) x Moles of KMnO4
= (5/2) x 0.000187 mol
= 0.0004675 mol
3) Determine the number of moles of FeSO4-7H2O:
From the balanced chemical equation, we know that 1 mole of KMnO4 oxidizes 1 mole of FeSO4-7H2O.
Moles of FeSO4-7H2O = Moles of Fe2+
= 0.0004675 mol
4) Calculate the mass of FeSO4-7H2O in the pill:
Mass of FeSO4-7H2O = Moles of FeSO4-7H2O x Molar Mass
= 0.0004675 mol x 278.03 g/mol
= 0.1299 g
5) Calculate the mass % of FeSO4-7H2O in the pill:
Mass % = (Mass of FeSO4-7H2O / Total mass of pill) x 100
= (0.1299 g / 0.4927 g) x 100
= 26.37%
Therefore, the mass percentage of FeSO4-7H2O in the iron supplement pill is approximately 26.37%.