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
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 = ?
The answer comes out to 418% but that doesn't seem to make sense. Is that the answer??
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.
To find the percentage of iron in the sample, we need to determine the moles of iron involved in the reaction.
First, let's find the moles of KMnO4 in the titration:
Moles of KMnO4 = concentration (M) × volume (L)
Moles of KMnO4 = 0.187 mol/L × 0.040 L
Moles of KMnO4 = 0.00748 mol
According to the balanced equation, the stoichiometric ratio between KMnO4 and Fe(II) is 1:5.
Therefore, moles of Fe(II) = 5 × moles of KMnO4
Moles of Fe(II) = 5 × 0.00748 mol
Moles of Fe(II) = 0.0374 mol
The molar mass of Fe is 55.85 g/mol.
The mass of iron in the sample = moles of Fe(II) × molar mass of Fe
The mass of iron in the sample = 0.0374 mol × 55.85 g/mol
The mass of iron in the sample = 2.088 g
Finally, we can calculate the percentage of iron in the sample:
Percentage of iron = (mass of iron in the sample / mass of the sample) × 100
Given that the sample mass is 0.500 g:
Percentage of iron = (2.088 g / 0.500 g) × 100
Percentage of iron = 417.6%
Therefore, the percentage of iron in the sample is 417.6%.