Asked by Julie
A .500g unknown sample is titrated with 40mL of .187M KMno4 by the following reaction:
5Fe{2+} + MnO4{-1} + 8H{+1} -> 5Fe{3+} + Mn{2+} + 4H2O.
What is the percentage of iron in the following sample
5Fe{2+} + MnO4{-1} + 8H{+1} -> 5Fe{3+} + Mn{2+} + 4H2O.
What is the percentage of iron in the following sample
Answers
Answered by
DrBob222
mols MnO4^- = M x L = ?
Using the coefficients in the balanced equation,convert mols MnO4^- to mols Fe.
Then g Fe = mols Fe x atomic mass Fe
Finally, %Fe = (g Fe/g sample)*100 = ?
Using the coefficients in the balanced equation,convert mols MnO4^- to mols Fe.
Then g Fe = mols Fe x atomic mass Fe
Finally, %Fe = (g Fe/g sample)*100 = ?
Answered by
Julie
The answer comes out to 418% but that doesn't seem to make sense. Is that the answer??
Answered by
DrBob222
No it doesn't make sense but your answer is correct for what you posted.
If M = 0.0187 (and 0.187M is higher than usual concentrations for KMNO4) it would be 41.8% which makes more sense. Check your post to make sure the numbers are right.
If M = 0.0187 (and 0.187M is higher than usual concentrations for KMNO4) it would be 41.8% which makes more sense. Check your post to make sure the numbers are right.
There are no AI answers yet. The ability to request AI answers is coming soon!