Question

What is the percentage of Fe2O3 in a sample of limonite ore if the iron from a 0.5000-g sample is reduced and titrated with 35.15 ml of a potassium dichromate solution of which 15.00 mL is equivalent in oxidizing power to 25.00 mL of a potassium permanganate solution which has an iron value of 0.004750 g?

Answer 1

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The first thing we will calculate is the iron value for the solution of potassium dichromate. Each mL of potassium permanganate solution titrates 0.004750 g of iron. So we use that relation to find the iron value of dichromate solution like this:

Iron value of dichromate = 0.004750 g/1 mL x 25 mL x 1 mL(dichromate) / 15 mL = 0.00792 g

Or we can find directly the amount of iron that the 15 mL represent:

15 mL of dichromate = 0.004750 g/1 mL x 25 mL = 0.11875 g

With this second value we can find out the amount of iron in the sample:

g of Fe = 0.11875 g / 15 mL x 35.15 mL = 0.2783 g

Remember that this is only iron and we are asked for the percentage of Fe₂O₃ in the sample. So we have to transform the g of iron in g of Fe₂O₃ using the molecular weights. Also remember that in the total weight of Fe₂O₃ there are two atomic weights of Fe:

g of Fe₂O₃ = 0.2783 g / 111.69 g of iron x 159.69 g of Fe₂O₃ = 0.3979 g

% of Fe₂O₃ = 0.3979 g Fe₂O₃ x 100 / 0.500 g sample = 79.58%

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