Question

Iron can be determined by its reaction with potassium permanganate by the reaction MnO4- + 5Fe + 2 + 8H + = Mn + 2 + 5Fe +3 + 4H20 A sample of 1.300g of steel was digested and diluted to complete 1L of The solution An aliquot of 25.00mL of said solution was taken and it was titrated with 12.20mL of 0.0100M KMnO4. The percentage of iron (Fw = 55.85) in the steel sample is?

Answer 1

no. of mole = molarity X volume of solution in liter

12.20 ml = 0.0122 liter

no. of KMnO₄ used for titration = 0.0100 X 0.0122 = 0.000122 mole

according to balanced reaction given in question 1 mole of KMnO₄ react with s mole of iron therefore to react with 0.000122 mole of KMnO₄ required iron = 5 X 0.000122 = 0.00061 mole

25 ml soluton contain 0.00061 mole iron then 1 L(1000ml) solution contain iron = 1000 X 0.00061/25 = 0.0244 mole of iron  

molar mass of iron = 55.85g/ mol

gm of compound = no.of mole X molar mass

gm of iron in sample = 0.0244 X 55.85 = 1.36 gm

according to your given KMnO₄ used for titration iron in sample = 1.36 gm

(may be u have given sample of steel value wrong or reaction wrong or KMnO₄ used value wrong)