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0.804 gm sample of iron ore containing only Fe and FeO was dissolved in acid. Iron oxidises into +2 states and it requires 117.20 ml of 0.112 N K2Cr2O7 solution for titration .Calculate the percentage of iron in the ore. one year ago

0.804 gm sample of iron ore containing only Fe and FeO was dissolved in acid. Iron oxidises into +2 states and it requires 117.20 ml of 0.112 N K2Cr2O7 solution for titration .Calculate the percentage of iron in the ore. one year ago

Grade:11

1 Answers

Rituraj Tiwari
askIITians Faculty 1792 Points
3 years ago
Since we know the equation that :-

6Fe2+ + Cr2O72− + 14H+ ->6 Fe3+ + 2Cr3+ + 7H2O.

Where iron is 6 mol and the Cr2O7 is of 1 mol.

We again know that the molarity of the K2Cr2O7 will be equal to the normality*n factor. Where for K2Cr2O7, we have the value of the n to be 6.

Hence, the molarity will be 0.112/6=0.018 M.

Now, the number of moles of K2Cr2O7 will be = 0.018×117.20/1000 =0.0021 moles.

Now, we get from the reaction that, 6 mol of Fe2+ which reacts with 1 mol of Cr2O72−. Therefore, 0.0021 mol of Cr2O72− will react with 6×0.0021=0.0131mol of the iron or Fe2+.

So, the mass of the Fe2+ will be 0.0131×55.85 = 0.7331 g.

Therefore, the mass of iron ore will be = 0.804 g% of the iron present in the ore = 0.73310.804×100=91.1%.

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