J.R. S. answered 01/13/18
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Do you know how to write oxidation/reduction half reactions? If not, you must learn.
(1) 2 MnO4- +5 H2C2O4 + 6 H+ =2 Mn2+ +10 CO2 + 8 H2O
(2) MnO4- + 5 Fe2+ + 8 H+ = Mn2+ + 5 Fe3+ + 4 H2O
Moles Fe2+ = 5 x 0.043 moles = 0.215 moles
(3) moles KMnO4 = 0.020 mole/L x 0.02715 L= 0.000543 moles MnO4- consumed
moles Fe2+ = 5 moles Fe2+ x 0.000543 moles KMnO4/mole Fe2+ = 0.00272 moles Fe2+ in sample
mass Fe = 0.00272 mol x 55.847 g/mol=0.152 g
% Fe = 0.152 g/1.630 g (x100%) = 9.33%
(2) MnO4- + 5 Fe2+ + 8 H+ = Mn2+ + 5 Fe3+ + 4 H2O
Moles Fe2+ = 5 x 0.043 moles = 0.215 moles
(3) moles KMnO4 = 0.020 mole/L x 0.02715 L= 0.000543 moles MnO4- consumed
moles Fe2+ = 5 moles Fe2+ x 0.000543 moles KMnO4/mole Fe2+ = 0.00272 moles Fe2+ in sample
mass Fe = 0.00272 mol x 55.847 g/mol=0.152 g
% Fe = 0.152 g/1.630 g (x100%) = 9.33%