# 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.

Mar 28, 2017

The ${\text{KMnO}}_{4}$ will react with 400 mg of $\text{Fe"_0.9"O}$.

#### Explanation:

Nonstoichiometric compounds

$\text{Fe"_0.9"O}$ is a nonstoichiometric compound.

Some of the $\text{Fe"^"2+}$ ions have been oxidized to $\text{Fe"^"3+}$.

Thus, to balance the charge, the compound contains two $\text{Fe"^"3+}$ ions for every three "missing" $\text{Fe"^"2+}$ ions.

There is still the same mass of $\text{Fe}$, 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

$\text{5FeO"color(white)(l) + "KMnO"_4 + "18H"^"+" → "5Fe"^"3+" + "Mn"^"2+" + "K"^"+" + "9H"_2"O}$

Calculations

${\text{Moles of KMnO"_4 color(white)(l)"used" = 0.010 color(red)(cancel(color(black)("L KMnO"_4))) × "0.1 mol MnO"_4/(1 color(red)(cancel(color(black)("L KMnO"_4)))) = "0.001 mol KMnO}}_{4}$

$\text{Moles of FeO reacted" = 0.001 color(red)(cancel(color(black)("mol KMnO"_4))) × "5 mol FeO"/(1 color(red)(cancel(color(black)("mol KMnO"_4)))) = "0.005 mol FeO}$

$\text{Moles of Fe"_0.9"O" = 0.005 color(red)(cancel(color(black)("mol FeO"))) × ("100 mol Fe"_0.9"O")/(90 color(red)(cancel(color(black)("mol FeO")))) = "0.0056 mol Fe"_0.9"O}$

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

Note: The answer can have only 1 significant figure, because that is all you gave for the molarity of the ${\text{KMnO}}_{4}$.