A 1.362 g sample of an iron ore that contained Fe_3O_4 was dissolved in acid and all the iron was reduced to Fe^2+. the solution was then acidified with H_2SO_4 and titrated with 39.42 mL of 0.0281 M KMnO_4 , which oxidized the iron to Fe^3+.

Question

A 1.362 g sample of an iron ore that contained Fe_3O_4 was dissolved in acid and all the iron was reduced to Fe^2+. the solution was then acidified with H_2SO_4 and titrated with 39.42 mL of 0.0281 M KMnO_4 , which oxidized the iron to Fe^3+.
Net ionic equation
5Fe^(2+)+MnO_4^-+8H^+→5Fe^(3+)+Mn^(2+)+4H_2O

a) what was the percentage by mass of iron in the ore?
b) what was the percentage by mass of Fe_3O_5 in the ore

DrBob222 · Answer

You have the balanced equation.
moles MnO4^- = M x L
Using the coefficients in the balanced equation, convert moles MnO4^- to moles Fe.
Now convert moles Fe to grams Fe. g = moles x molar mass.

%Fe = (mass Fe/mass sample)*100 = ??

Convert mass Fe to mass Fe3O4. (I assume that is Fe3O4 and not Fe3O5.)
%Fe3O4 = (mass Fe3O4/mass sample)*100 = ??