Answer :
Answer:
% Fe in the sample = 22.0%
Explanation:
The redox reaction between iron and permanganate is:
[tex]MnO_{4}^{-}+5Fe^{2+}+8H^{+}\rightarrow Mn^{2+}+5Fe^{3+}+4H2O[/tex]
Moles of permanganate required is:
[tex]molarity*volume = 0.0194 mols/L*0.02022L = 0.000392\ moles[/tex]
Based on the reaction stoichiometry:
5 moles of iron requires 1 mole of permanganate solution
Therefore, 0.000392 mols of permanganate would need:
[tex]=\frac{0.000392\ mols\ permanganate*5\ mols\ iron}{1\ mol\ permanganate} =0.00196\ mols[/tex]
Mol mass of iron (Fe) = 55.845 g/mol
Mass of iron present is:
[tex]=moles*mol.mass = 0.00196mols*55.845g/mol=0.109 g[/tex][tex]Percent\ iron = \frac{Mass(iron)}{Mass(sample)} *100\\\\Percent\ iron =\frac{0.109}{0.495} *100=22.0[/tex]