[necroramo]

The iron content of iron ore can be determined by titration with standard KMnO4 solution. The iron ore is dissolved in excess HCL, and the iron reduced to Fe+2 ions. This solution is then titrated with KMnO4 solution, producing Fe+3 and Mn+2 ions in acidic solution. If it requires 45.95 mL of .205 M KMnO4 to titrate a solution made with .6128 g of iron ore, what percent of the ore was iron?

Anyway,

I got the initial imbalanced equation
Fe+2 + MnO4- --> Fe+3 + Mn+2
Then
Fe+2 --> Fe+3 + e-
8H+ + 5e- + MnO4- --> Mn+2 + 4H20

combining them makes
8H+ + 5Fe+2 + MnO4 --> 5Fe+3 + Mn+2 + 4H20

then, however, it becomes mole conversion time.

I got .00860 mol of KMnO4 and thus MnO4-, converting to .0430 mol of Fe.... which is 2.4 g, much more than what was in the ore.

Where am I screwing this up?

[Astrokel]

hello!

I got .00860 mol of KMnO4

Check again, but i have tried working with the correct amount of KMnO4, but it doesn't seem to work, unless you got your concentration of KMnO4 wrong, 0.205 or 0.0205?

hope this helps some...

[Borek]

I got .00860 mol of KMnO4 and thus MnO4-

I got 0.00942 (0.04595*0.205)

converting to .0430 mol of Fe

0.0471

which is 2.4 g

2.63 g

much more than what was in the ore.

agreed

Where am I screwing this up?

Astrokel had a good point.

Full view from stoichiometry calculator won't fit the screen, only simplified version.

[necroramo]

It definitely says .205 M.

[Borek]

If it says 0.205M, 45.95 mL and 0.6128 g, then it is definitely wrong.