Question

A sample of 1.55 g of iron ore is dissolved in an acid solution in which the iron is converted into
Fe2+. The solution formed is then titrated with KMnO4 which oxidises Fe2+ to Fe3+ while the MnO4-
ions are reduced to Mn2+ ions. 92.95 mL of 0.020 M KMnO4 is required for the titration to reach
the equivalence point.
a) Write the balanced equation for the titration.
b) Calculate the percentage of iron in the sample.​

Answer 1

Answer:a) 8H2SO4 + 2KMnO4 + 10FeSO4 → 5Fe2(SO4)3 + 8H2O + 2MnSO4 + K2SO4

B) nKMno4 = 0.001859 mol

=>nFeso4=0.009295 mol

nFe= 0.009295 mol

mFe=0.52052 g

=>percentage of iron in the sample: 33.5819%

Explanation:a) 8H2SO4 + 2KMnO4 + 10FeSO4 → 5Fe2(SO4)3 + 8H2O + 2MnSO4 + K2SO4

B) nKMno4 = 0.001859 mol

=>nFeso4=0.009295 mol

nFe= 0.009295 mol

mFe=0.52052 g

Answer 2

The reaction is a redox reaction.

The percentage of Fe^2+ is  34.3%

The balanced overall redox reaction equation is:

$MnO4^- + 8H^+ + 5Fe^2+ ----> Mn^2+ + 4H2O + 5Fe^3+$

Number of moles of permanganate =  0.020 M  * 92.95/1000 L = 0.0019 moles

From the reaction equation:

5 moles of $Fe^2+$reacts with 1 mole of permanganate

x moles of $Fe^2+$ reacts with 0.0019 moles of permanganate

x = 5 * 0.0019 /1 = 0.0095 moles of $Fe^2+$

Mass of $Fe^2+$ = 0.0095 moles of $Fe^2+$ * 56 g/mol = 0.532 g

Percentage of iron = 0.532 g/1.55 g  * 100  = 34.3%

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