posted by avant on .
This is a hard one. I'll do my best to try and translate =)
A sour liquid mixture (many H+) contains Fe2+ ions and is being titrated with 0,0202M MnO4-. At the end of the titration, the amount of MnO4- that had been used up was 20cm3. How much was the mass of Fe2+ in the first mixture?
I've counted out the amount of mol for the MnO4-, and then used that against the mol/g of Fe2+, but it doesn't add up.
mols MnO4^- = 0.0202 M x 0.020 L = ??Next, write the equation for the titration rxn.
MnO4^- + Fe^+2 ==> Mn^+2 + Fe^+3
You can break it down into the half reactions and write
MnO4^- ==> Mn^+2 and
Fe^+2 ==> Fe^+3
then balance but the end result will be MnO4^- + 5Fe^+2 ==> 5Fe^+3 + Mn^+2
So mols Fe^+2= ??mols MnO4^-(from above) x (5 mols Fe^+2/1 mol MnO4^-) = xx mols Fe^+2.
g Fe^+2 = xx mols Fe^+2 x molar mass Fe.
I hope this helps. If you get stuck and need further assistance, please post your work. It will help me know where you are having a problem.