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#.
