Answer:
% Iron in the ore = 2.57%
Explanation:
The redox reaction involving the oxidation of Fe2+ to Fe3+ in the presence of Na2Cr2O7 is:
[tex]6Fe^{2+} + Cr_{2}O_{7}^{2-}+14H^{+}\rightarrow 6Fe^{3+}+2Cr^{3+}+7H_{2}O[/tex]
Based on the reaction stoichiometry:
6 moles of Fe2+ requires with 1 mole of Cr2O7^2-
Moles of Cr2O7^2- required in the titration:
[tex]= Molarity*Volume = 0.0100 moles/L * 0.01917 L = 0.0001917 moles[/tex]
Therefore, moles of Fe2+ present is:
[tex]= 6*0.0001917 = 0.00115 moles[/tex]
Atomic mass of Fe = 55.85 g/mol
Mass of Fe present in the ore is:
[tex]= Moles*atomic.wt = 0.00115 moles *55.85 g/mol = 0.06423 g[/tex]
%Fe in the ore is: [tex]=\frac{Mass(Fe)}{Mass(Ore)}*100 = \frac{0.06423}{2.5000} *100 = 2.57%[/tex]
