Asked by Sam
an 0.5031 gram sample of iron ore is determined through redox titration using K2Cr2O7. The sample was dissolved in HCl and all the iron was converted to the 2+ state. Titration to the endpoint required 24.34 mL of 0.01887 mol/L K2Cr2O7. Calculate the amount of Fe2O3 as a percent of the original sample.
Answers
Answered by
DrBob222
Fe2O3 ==> 2Fe^3+ and reduce to Fe^2+.
6Fe^2+ + Cr2O7^2- ==> 6Fe^3+ + 2Cr^3+
I balanced only the redox portion of the equation which is sufficient for what we want to do.
mols Cr2O7^2- = M x L = ?
mols Fe = 6 x that (look at the coefficients in the equation above).
g Fe = mols Fe x atomic mass Fe = ?
%Fe = (g Fe/mass sample)*100 = ?
then convert %Fe to %Fe2O3 this way.
%Fe2O3 = %Fe x (molar mass Fe2O3/2*atomic mass Fe) = ?
6Fe^2+ + Cr2O7^2- ==> 6Fe^3+ + 2Cr^3+
I balanced only the redox portion of the equation which is sufficient for what we want to do.
mols Cr2O7^2- = M x L = ?
mols Fe = 6 x that (look at the coefficients in the equation above).
g Fe = mols Fe x atomic mass Fe = ?
%Fe = (g Fe/mass sample)*100 = ?
then convert %Fe to %Fe2O3 this way.
%Fe2O3 = %Fe x (molar mass Fe2O3/2*atomic mass Fe) = ?
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