Asked by Child
A 0.020 M KMnO4 solution was used to determine
the percent iron in a sample containing
iron(II).
MnO−
4 + 5 Fe2+ + 8 H+ →
Mn+2 + 5 Fe3+ + 4 H2O
It took 25.0 mL of the KMnO4 to completely
react with all the iron(II) in the 0.500 g sample.
What was the percent iron(II) in the
sample?
the percent iron in a sample containing
iron(II).
MnO−
4 + 5 Fe2+ + 8 H+ →
Mn+2 + 5 Fe3+ + 4 H2O
It took 25.0 mL of the KMnO4 to completely
react with all the iron(II) in the 0.500 g sample.
What was the percent iron(II) in the
sample?
Answers
Answered by
DrBob222
mols KMnO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols KMnO4 to mols Fe(II).
Convert mols Fe(II) to grams Fe. g = mols x atomic mass Fe.
%Fe = (mass Fe/mass sample)*100 = ?
Using the coefficients in the balanced equation, convert mols KMnO4 to mols Fe(II).
Convert mols Fe(II) to grams Fe. g = mols x atomic mass Fe.
%Fe = (mass Fe/mass sample)*100 = ?
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