Question

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.

Answer 1

Answer:

91.1%.

Explanation:

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%.