Question

A 0.495 g sample of iron containing salt required 20.22 mL of 0.0194 M permanganate solution to reach the endpointof a titration. What is the percent of iron (55.845 g/mol) in the salt

Answer 1

Answer:

% Fe in the sample = 22.0%

Explanation:

The redox reaction between iron and  permanganate is:

$MnO_{4}^{-}+5Fe^{2+}+8H^{+}\rightarrow Mn^{2+}+5Fe^{3+}+4H2O$

Moles of permanganate required is:

$molarity*volume = 0.0194 mols/L*0.02022L = 0.000392\ moles$

Based on the reaction stoichiometry:

5 moles of iron requires 1 mole of permanganate solution

Therefore, 0.000392 mols of permanganate would need:

$=\frac{0.000392\ mols\ permanganate*5\ mols\ iron}{1\ mol\ permanganate} =0.00196\ mols$

Mol mass of iron (Fe) = 55.845 g/mol

Mass of iron present is:

$=moles*mol.mass = 0.00196mols*55.845g/mol=0.109 g$$Percent\ iron = \frac{Mass(iron)}{Mass(sample)} *100\\\\Percent\ iron =\frac{0.109}{0.495} *100=22.0$