A 0.8040-g sample of an iron ore is dissolved in acid. The iron is then reduced
ID: 954276 • Letter: A
Question
A 0.8040-g sample of an iron ore is dissolved in acid. The iron is then reduced to Fe^2+ and titrated with 47.22 mL of 0.02242 M KMnO_4 solution. calculate the results of this analysis in terms of (a) % Fe (55.847 g/mol) and (b) % Fe_3O_4 (231.54 g/mol). The reaction of the analyte with the reagemt is described by the equation MnO_4 + 5Fe^2+ + 8H^+ rightarrow Mn^2+ + 5Fe^3+ + 4H_2O (a) stoichiometric ratio = 5 mmol Fe^2+/1 mmol KMnO_4 amount KMnO_4 = 47.22 mL, KMnO_4 times 0.02242 mmol KMnO_4/mL KMnO_4 The mass of Fe^2+ is then given by mass Fe^2+ = (47.22 times 0.02242 times 5) mmol Fe^2+ times 0.055847 gFe^2+/mmol Fe^2+ The percent Fe^2+ is % Fe^2+ = (47.22 times 0.02242 times 5 times 0.055847) g Fe^2+/0.8040 g sample times 100 % = 36.77% (b) To determine the correct stoichiometric ratio, we note that 5 Fe^2+ 1 MnO_4 Therefore, 5Fe_3O_4 15Fe^2+ 3MnO_4 and stoichiometric ratio = - 5 mmo1 Fe_3O_4/3 mmol KMnO_4 As in part (a), amount KMnO, = 47.22 mL KMnO_4 times 0.02242 mmol KMnO_4/mL KMnO_4Explanation / Answer
Answer:
KMnO4 + 5 Fe{2+} + 8 H{+} Mn{2+} + 5 Fe{3+} + 4 H2O + K{+}
a.)
(0.04722 L) x (0.02242 mol/L KMnO4) x (5 mol Fe / 1 mol KMnO4) x
(55.8452 g Fe/mol) / (0.8040 g) = 0.3677 = 36.77% Fe
b.)
(0.04722 L) x (0.02242 mol/L KMnO4) x (5 mol Fe / 1 mol KMnO4) x
(1 mol Fe3O4 / 3 mol Fe) x (231.5333 g Fe3O4/mol) / (0.8040 g) = 0.5081 = 50.81% Fe3O4
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