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?
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 = ?
To determine the percent iron(II) in the sample, we need to calculate the moles of KMnO4 used and the moles of Fe2+ present in the sample. By comparing these two values, we can calculate the percent iron(II) in the sample.
Let's break down the steps to get the answer:
Step 1: Calculate the moles of KMnO4 used.
Given that the solution has a concentration of 0.020 M (moles per liter) and the volume used is 25.0 mL (0.025 L), we can use the formula:
Moles = Concentration × Volume
Moles of KMnO4 = 0.020 M × 0.025 L
Step 2: Calculate the moles of Fe2+ in the sample.
Using the balanced chemical equation provided:
MnO−4 + 5 Fe2+ + 8 H+ → Mn+2 + 5 Fe3+ + 4 H2O
We can observe that it takes 5 moles of Fe2+ to react with 1 mole of KMnO4.
Therefore, Moles of Fe2+ = (Moles of KMnO4) × 5
Step 3: Calculate the weight of Fe2+ in the sample.
Given that the molar mass of Fe2+ is 55.85 g/mol, we can use the formula:
Weight = Moles × Molar Mass
Weight of Fe2+ = (Moles of Fe2+) × (Molar mass of Fe2+)
Step 4: Calculate the percent of Fe2+ in the sample.
The weight of Fe2+ in the sample is given as 0.500 g.
Using the formula:
Percent = (Weight of Fe2+ / Sample weight) × 100
Percent Fe2+ in the sample = (Weight of Fe2+ / 0.500 g) × 100
By plugging in the values calculated in the previous steps, you can find the percent iron(II) in the sample.