Asked by Yoojin Lee
Calculate the percent of iron in a 50.0 gram sample containing iron if 45.6 mL of 2.5 M KMnO4 titrant was used. can you solve this?
Answers
Answered by
DrBob222
Did you put the Fe in solution and reduce it to Fe^2+ for titration with KMnO4. If so, then
MnO4^- + 5Fe^2+ + H^+ ==> Mn^2+ + 5Fe^3+
Note: This is not the entire equation; therefore, it isn't balanced. However, the part that matters (the redox part) IS balanced. You may finish the equation if needed.
mols MnO4^- = M x L = ?
Using the coefficients in the above equation convert mols MnO4^- to mols Fe. Then g Fe = mols Fe x atomic mass Fe.
Finally, %Fe = (g Fe/g sample)*100 = ?
MnO4^- + 5Fe^2+ + H^+ ==> Mn^2+ + 5Fe^3+
Note: This is not the entire equation; therefore, it isn't balanced. However, the part that matters (the redox part) IS balanced. You may finish the equation if needed.
mols MnO4^- = M x L = ?
Using the coefficients in the above equation convert mols MnO4^- to mols Fe. Then g Fe = mols Fe x atomic mass Fe.
Finally, %Fe = (g Fe/g sample)*100 = ?
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