The amount of Fe in a 2.0008 g sample of an iron ore was determined by a redox titration with K2Cr2O7. The sample was dissolved in HCl and the iron brought to a +2 oxidation state. Titration required 26.48 mL of 0.034 M K2Cr2O7. Calculate the iron content as %w/w Fe2O3.
Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
hint: Remember you are looking for Fe2O3 (2 mols of Fe)
mols Cr2O7^2- = M x L = ?
Convert moles Cr2O7^2- to mol Fe.
mols Fe = mols Cr2O7^2- x (6 mol Fe/1 mol Cr2O7^2-) = ? (from the equation)
Convert to moles Fe2O3.
mol Fe x (1 mol Fe2O3/2 moles Fe) = ?
g Fe2O3 = moles Fe2O3 x molar mass = ?
%Fe2O3 = (mass Fe2O3/mass sample)*100 = ?
To calculate the iron content as %w/w Fe2O3, we need to determine the amount of Fe2O3 present in the 2.0008 g sample of iron ore.
1. First, we need to find the moles of K2Cr2O7 used in the titration.
Moles of K2Cr2O7 = concentration (M) × volume (L)
Moles of K2Cr2O7 = 0.034 M × 0.02648 L
Moles of K2Cr2O7 = 0.00089792 mol
2. According to the balanced redox equation, it takes 6 moles of Fe2+ to react with 1 mole of K2Cr2O7.
Therefore, the moles of Fe2+ present in the sample can be calculated as:
Moles of Fe2+ = 6 × Moles of K2Cr2O7
Moles of Fe2+ = 6 × 0.00089792 mol
Moles of Fe2+ = 0.00538752 mol
3. From the balanced equation, we know that for every 6 moles of Fe2+ there are 6 moles of Fe3+.
Therefore, the moles of Fe3+ present in the sample are the same as the moles of Fe2+, which is 0.00538752 mol.
4. Now, we need to calculate the moles of Fe2O3.
From the balanced equation, we know that for every 2 moles of Fe3+, there is 1 mole of Fe2O3.
Moles of Fe2O3 = Moles of Fe3+ / 2
Moles of Fe2O3 = 0.00538752 mol / 2
Moles of Fe2O3 = 0.00269376 mol
5. Finally, we can calculate the iron content as %w/w Fe2O3.
Mass of Fe2O3 = Moles of Fe2O3 × molar mass of Fe2O3
Mass of Fe2O3 = 0.00269376 mol × (2 × atomic mass of Fe + 3 × atomic mass of O)
Mass of Fe2O3 = 0.00269376 mol × (2 × 55.845 g/mol + 3 × 16.00 g/mol)
Mass of Fe2O3 = 0.00269376 mol × 159.688 g/mol
Mass of Fe2O3 = 0.4296 g
%w/w Fe2O3 = (mass of Fe2O3 / mass of sample) × 100
%w/w Fe2O3 = (0.4296 g / 2.0008 g) × 100
%w/w Fe2O3 = 21.472%
Therefore, the iron content as %w/w Fe2O3 in the iron ore sample is approximately 21.472%.