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Chemistry Help!?!?!?!?
A sample of a salt that contains iron (II) ions, with a mass of 9.252 g, was dissolved in distiller water and made up to the mark in a 250.0 mL volumetric flask. 25.00 mL aliquot are pipetted from the stock solution, acidified, and titration with 0.02500 M potassium manganate(VII) solution. The manganate(VII) solution acts as an oxidizing agent in acid solution.
The volumes of manganate(VII) solution required to reach the end-point in three separate experiments were found to be, 20.95 mL, 20.75 mL and 20.90 mL respectively.
Use these data to calculate the % of iron in the salt.
1 Answer
- Roger the MoleLv 77 years agoFavorite Answer
5 Fe{2+} + Mn{7+} → 5 Fe{3+} + Mn{2+}
(20.95 mL + 20.75 mL + 20.90 mL) / 3 = 20.867 mL on average
(0.020867 L) x (0.02500 mol/L Mn{7+}) x (5 mol Fe / 1 mol Mn{7+}) x (55.8452 g Fe/mol) x
(250.0 mL / 25.00 mL) / (9.252 g) = 0.1574 = 15.74% Fe
