titration problem Bryce decided to use potassium permanganate to determine the p

Question

titration problem

Bryce decided to use potassium permanganate to determine the percentage of iron (Fe, At. Wt. = 55.845) in an ore sample. His procedure included dissolving the iron with HCl and then converting all the iron in the ore to Fe2+ using several reagents. Once completing the conversion, Bryce titrated the resulting Fe2+ with Mn04-: Fe2+ (aq) + Mn04- (aq) rightarrow Fe3+ (aq) + Mn2+ (aq) (unbalanced) His sample of the original ore was 3.852 g and was processed for titration in a total volume of 150.0 mL of solution. A 50.0 mL aliquot of this solution was used for the titration with a 0.0.0512 M KMn04 solution. A total of 46.2 mL was required to obtain the light purple-pink endpoint. What is the percent of iron in the ore sample? What if any, recommendations

Explanation / Answer

Fe moles initially = 3.852/55.845 = 0.0069

vol = 150 ml , M1 = 0.0069/0.15 = 0.46

now 50 ml taken out

moles of Fe2+ = 0.46 x 0.05 = 0.023

moles of MnO4- used = 0.0512 x0.0462 = 0.002365

moles of Fe2+ stochiometrically reacted = 0.002365 x 5 =0.011827

moles of Fe2+ (impure sol ) used = 0.023

hence Fe2+ in sample = (0.011827/0.023) x100 = 51.42 %

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