[bnist_nazrul]

an iron ore is analyzed for iron content by dissolving in acid ,converting iron to Fe, and then titrating with standard 0.0150M K2Cr2O7 solution. if 35.6 ml is reqires to titrate the iron ina 1.68g ore sample, expressed as percent Fe2O3?

[mikasaur]

Hello and welcome to the forums! Before we can help you though, you must show your attempt at completing the problem. It's all described in the rules here.

That said, I think you might have left out some words in your question. As is it doesn't really make sense.

[Borek]

converting iron to Fe

You don't convert iron to iron, you have ignored one important fact about the conversion.

[bnist_nazrul]

Potassium dichromate converts Fe(II) to Fe(III) (redox reaction)

Dichromate reacts with iron(II) as follows:
Cr2O72- + 6 Fe2+ + 14H+ → 2Cr3+ + 6 Fe3+ + 7H2O

1 mole of potassium dichromate reacts with 6 moles of FeO.

35.6 ml of 0.015 M dichromate solution contains 35.6 / 1000 * 0.015 = 0.000534 M dichromate

This would react with 6 * 0.000534 = 0.003204 M of Fe2+

0.003204 M of FeO is 72 * 0.003204 = 0.231 g


bUT answer is 15.3%..i confused..

[Borek]

This would react with 6 * 0.000534 = 0.003204 M of Fe2+

Almost OK so far, just it is 0.003205 moles, not M (M stands for molar concentration).

0.003204 M of FeO is 72 * 0.003204 = 0.231 g

Why FeO? You are asked to express the result as percentage of Fe₂O₃.

In how many moles of Fe₂O₃ are 0.003204 moles of Fe present?