Back titration question, sac on Monday please help!

[ganesh]

Hey guys! Super stuck on this question, I think I've missed a step or something, I keep getting the answer of n=0.8, however the correct answer is n=3! PLEASE HELP.

A 1.0101 gram sample of Fe(OH)n was mixed with 20.00 mL of 2.000 M HCl and enough water added to make 200.0 mL of solution A.
Fe(OH)n(s) + nHCl(aq) FeCln(aq) + nH2O(l)
A 25.00 mL aliquot of solution A was taken and titrated with 14.56 mL of 0.1000 M KOH.
KOH(aq) + HCl(aq) H2O(l) + KCl(aq)
What is the value of n?

[Mhysa]

 0.01456 L * 0.1000 mol/L * 200/25 = 0.011648 moles HCl that didn't react.
0.020 L * 2.000 mol/L = 0.0400 original mol HCl.
0.0400 - 0.011648 = 0.028352 moles HCl reacted.
Assuming n = 3, then 1.0101 g / 106.867 g/mol =0.00945 mol.
Ratio = 0.02845 / 0.00945 = 3.00.
I had to assume n was 3 to get the molar mass to make sense of the mass.
The Fe(OH)n / HCl ratio was 3 based on this assumption.

[nerdgasm]

I think there might be a way to deduce that the value of n is 3, but it does require some algebra:

After working out that 0.028352 moles of HCl reacted, we then turn to the original equation we wrote: Fe(OH)n(s) + nHCl(aq) ->  FeCln(aq) + nH2O(l). From this, we can see that the molar ratio between Fe(OH)n and HCl is 1:n.  Therefore, since 0.028352 moles of HCl reacted, it follows that 0.028352/n moles of Fe(OH)n reacted. Also, we assume that all 1.0101g of Fe(OH)n reacted, because we had excess HCl (which was later titrated against KOH). Finally, note that the molar mass of Fe(OH)n will be 55.85 + n(17.008).

We can now write the following statement:
0.028352/n moles of a compound with molar mass 55.85 + n(17.008) has a mass of 1.0101g.  Using the formula Mass = Molar Mass * No. of moles, we get: