Question

Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 M Na2Cr2O7 is required in the titration. What percentage of the ore sample was iron?

Answer 1

% Iron in the ore = 2.57%

Step-by-step explanation:

The redox reaction involving the oxidation of Fe2+ to Fe3+ in the presence of Na2Cr2O7 is:

`6Fe^(2+) + Cr_(2)O_(7)^(2-)+14H^(+)\rightarrow 6Fe^(3+)+2Cr^(3+)+7H_(2)O`

Based on the reaction stoichiometry:

6 moles of Fe2+ requires with 1 mole of Cr2O7^2-

Moles of Cr2O7^2- required in the titration:

`= Molarity*Volume = 0.0100 moles/L * 0.01917 L = 0.0001917 moles`

Therefore, moles of Fe2+ present is:

`= 6*0.0001917 = 0.00115 moles`

Atomic mass of Fe = 55.85 g/mol

Mass of Fe present in the ore is:

`= Moles*atomic.wt = 0.00115 moles *55.85 g/mol = 0.06423 g`

%Fe in the ore is:
`=(Mass(Fe))/(Mass(Ore))*100 = (0.06423)/(2.5000) *100 = 2.57%`