posted by Linda .
In a previous experiment, 0.800 g of the green complex iron salt was placed in an Erlenmeyer flask and heated with concentrated sulfuric acid until all oxalate ion was destroyed (CO2, SO2 and H2O are formed). Ferric ion (Fe3+) remained in solution and was reduced to Fe2+ by reaction with Al wire. The ferrous ion (Fe2+) was then titrated with KMnO4 solution according to equation (2) and 32.60 mL of 0.0100 M permanganate solution was required to reach the endpoint. The blank required 0.03 mL of the permanganate solution. Find the mass % of iron in the complex salt.
1. Write the balanced equation for KMnO4 + Fe+2 ==>
2. M KMnO4 x L KMnO4 = mols KMnO4.
3. Convert mols KMnO4 to mols iron using the coefficients in the balanced equation.
4. mols Fe x atomic mass Fe = grams Fe.
5. (g Fe/mass sample)*100 = percent Fe.
Note: L of KMnO4 must be corrected for the blank. 32.60 mL - 0.03 mL = KMnO4 used to titrate the iron.
Post your work if you get stuck.
Is this correct? I did not put the charges
1MnO4 + 8H + 5Fe2->1Mn+ 4H20 + 5Fe
Volume of KMnO4= 32.6-.03=32.57 mL
32.57mL*(1L/1000mL)*(8.58x10^-3M)*(5mol Fe/1 mol KMnO4)*(55.85g Fe/mol)
I thought the problem stated the molarity as 0.01 M. Where did you find the 8.58 x 10^-4 M?
Substitute 0.01 for that "strange" number and that will be the grams of iron.
Even though it's been years, 8.58 x 10^-4 M would be the standardized solution of KMnO4. This is usually given by the lab instructor at the beginning of the experiment.