potassium permanganate was used to determine the percentage of iron (Fe, At weig

Question

potassium permanganate was used to determine the percentage of iron (Fe, At weight: 55.845? in an ore sample. the procedure included dissolving the iron wiht HCl and then converting all the iron in the ore to Fe2+ using several reagents. Once completing the conversion, titrated the following resulting Fe2+ with MnO4-:

Fe2+ + MnO4- --> Fe3+ + Mn2+ ( unbalanced)

The sample of the original ore was 3.852g and was processed for titration in the a total volume of 150.0 ml solution. a 50.0 ml aliquot of this solution was used for the titration with a .0512 M KMnO4 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?

Explanation / Answer

Fe moles initially = 3.852/55.845 = 0.0069

vol = 150 ml , M1 = 0.0069/0.15 = 0.046

now 50 ml taken out

moles of Fe2+ = 0.046 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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