How many milli grams of #"Fe"_0.9"O"# reacts completely with 10 mL 0.1 M KMnO4 solution in acidic conditions. (Fe = 56)? thanking you in anticipation.

1 Answer
Mar 28, 2017

Answer:

The #"KMnO"_4# will react with 400 mg of #"Fe"_0.9"O"#.

Explanation:

Nonstoichiometric compounds

#"Fe"_0.9"O"# is a nonstoichiometric compound.

Some of the #"Fe"^"2+"# ions have been oxidized to #"Fe"^"3+"#.

Thus, to balance the charge, the compound contains two #"Fe"^"3+"# ions for every three "missing" #"Fe"^"2+"# ions.

There is still the same mass of #"Fe"#, but only 90 % of it is in the +2 oxidation state.

Stoichiometry

The system will behave as if the reaction goes only to 90 % completion.

The equation for the reaction is

#"5FeO"color(white)(l) + "KMnO"_4 + "18H"^"+" → "5Fe"^"3+" + "Mn"^"2+" + "K"^"+" + "9H"_2"O"#

Calculations

#"Moles of KMnO"_4 color(white)(l)"used" = 0.010 color(red)(cancel(color(black)("L KMnO"_4))) × "0.1 mol MnO"_4/(1 color(red)(cancel(color(black)("L KMnO"_4)))) = "0.001 mol KMnO"_4#

#"Moles of FeO reacted" = 0.001 color(red)(cancel(color(black)("mol KMnO"_4))) × "5 mol FeO"/(1 color(red)(cancel(color(black)("mol KMnO"_4)))) = "0.005 mol FeO"#

#"Moles of Fe"_0.9"O" = 0.005 color(red)(cancel(color(black)("mol FeO"))) × ("100 mol Fe"_0.9"O")/(90 color(red)(cancel(color(black)("mol FeO")))) = "0.0056 mol Fe"_0.9"O"#

#"Mass of Fe"_0.9"O" = 0.0056 color(red)(cancel(color(black)("mol Fe"_0.9"O"))) × ("71.84 g Fe"_0.9"O")/(1 color(red)(cancel(color(black)("mol Fe"_0.9"O")))) = "0.4 g Fe"_0.9"O" = "400 mg Fe"_0.9"O"#

Note: The answer can have only 1 significant figure, because that is all you gave for the molarity of the #"KMnO"_4#.