General

Chemistry Redox Titration

*NVM. I'm dumb.. I put 2 electrons for the iron for some reason when it was suppose to be 1..
Need help once again D:. How do you do this redox titration question?
A student used an acidified 6.31x10^-2 mol/L KMnO4(aq) to titrate 25.0mL samples of Fe2+(aq) solution of unknown concentration. In the reactions, the Fe2+(aq) ion was oxidized to the Fe3+(aq) ion. The student completed five trials and summarized a data in a table. Er.. I can't really type the table, so I'll just give you the avg of the 3 closest titrations (16.3mL).
Ok so here's what I did (for the first few parts it should be correct)
K+, Mno4-, H+, H2O
so the SOA is Mno4- and H+ (note I'm not going to bother with the states, because that's not the problem here lol)
(x2) Mno4- + (8 H+) + 5e- ----> Mn2+ + 4H2O
(x5) Fe2+----> (Fe3+) + 2e-
= (2Mno4)- + (16 H+) + (5 Fe2+) ---> (2 Mn2+) + (5 Fe3+) + (8H2O)
(put it in brackets, because it looks a little confusing to me lol)
n= C x v = 0.0631mol/L x 0.163L <--(avg of the KMno4)
= 0.00102853 mol
(ok here's the part I'm doing wrong... I think I just forgot/fail it)
I'm trying to convert the moles of the Mno4- to Fe2+?
so...
0.00102853mol x 2/5? or 5/2? either way I'm still doing it wrong... Here's my wrong thinking process: There's 2 Mno4- in the balanced equation and 5 Fe2+
and I would get 0.002571325
C= n/v = 0.0025...mol/ 0.025L
= 0.102853mol/L
The answer is 0.206mol/L (rounded)... so I'm not suppose to divide it by two in the part that I got wrong..? why? D:

And once again Thanks again to all who help
@Burning considering you're one of the people who I know is really good at this stuff , if you can do it would be SUPER greatly appreciated

NVM. I'm dumb.. I put 2 electrons for the iron for some reason when it was suppose to be 1..