100 003
Assignments Done
98.3%
Successfully Done
In June 2021

Answer to Question #202765 in Chemistry for Ra Rayo

Question #202765

The amount of iron in a meteorite is determined by a redox titration using KMnO4 as the titrant. A 0.4185-g sample is dissolved in acid and the liberated Fe3+ quantitatively reduced to Fe2+ using a Walden reductor. Titrating with 0.02500 M KMnO4 requires 41.27 mL to reach the endpoint. Determine the %w/w Fe2O3 in the sample of meteorite.

1
2021-06-04T16:11:05-0400

Solution:

Balanced chemical equation for the titration reaction:

5Fe2+ + MnO4 + 8H+ → 5Fe3+ + Mn2+ + 4H2O

By stoichiometry, n(Fe2+)/5 = n(MnO4)

n(MnO4) = C(MnO4) × V(MnO4) = 0.02500 M × 0.04127 L = 0.00103175 mol

Therefore,

n(Fe2+) = 5 × n(MnO4) = 5 × 0.00103175 mol = 0.00515875 mol

Fe2O3 → 2Fe3+ → 2Fe2+

According to the scheme above: n(Fe2O3) = n(Fe2+)/2

n(Fe2O3) = n(Fe2+) / 2 = 0.00515875 mol / 2 = 0.002579 mol

The molar mass of Fe2O3 is 159.69 g/mol.

Therefore,

(0.002579 mol Fe2O3) × (159.69 g Fe2O3 / 1 mol Fe2O3) = 0.4119 g Fe2O3

Mass of Fe2O3 = 0.4119 g

%w/wFe2O3 = (mass of Fe2O3 / mass of sample) × 100%

%w/wFe2O3 = (0.4119 g / 0.4185 g) × 100% = 98.42%

%w/wFe2O3 = 98.42%

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!