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

% Fe in the sample = 22.0%

Step-by-step 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:

`=(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 = (Mass(iron))/(Mass(sample)) *100\\\\Percent\ iron =(0.109)/(0.495) *100=22.0`