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.

Question

Cameron Andrews · Accepted Answer

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

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#.

Serenity Allen · Answer

undefined To solve this problem, we first need to determine the moles of $ \text{KMnO}_4 $ used in the reaction:

$$
\text{Moles of KMnO}_4 = \text{Volume of solution (L)} \times \text{Molarity (mol/L)}
$$

$$
\text{Moles of KMnO}_4 = 10 \times 0.1 = 1 \text{ mmol}
$$

Since $ \text{Fe}_2\text{O}_3 $ reacts with $ \text{KMnO}_4 $ in acidic conditions according to the following equation:

$$
5\text{Fe}^{2+} + 16\text{H}^+ + 2\text{MnO}_4^- \rightarrow 5\text{Fe}^{3+} + 8\text{H}_2\text{O} + 2\text{Mn}^{2+}
$$

We can see that 2 moles of $ \text{KMnO}_4 $ react with 5 moles of $ \text{Fe}_2\text{O}_3 $. Therefore, 1 mmol of $ \text{KMnO}_4 $ will react with $ \frac{5}{2} $ mmol of $ \text{Fe}_2\text{O}_3 $.

Now, let's calculate the mass of $ \text{Fe}_2\text{O}_3 $ needed:

$$
\text{Moles of Fe}_2\text{O}_3 = \frac{5}{2} = 2.5 \text{ mmol}
$$

$$
\text{Mass of Fe}_2\text{O}_3 = \text{Moles} \times \text{Molar mass} = 2.5 \times 56 = 140 \text{ mg}
$$

So, 140 milligrams of $ \text{Fe}_2\text{O}_3 $ will react completely with 10 mL of 0.1 M $ \text{KMnO}_4 $ solution in acidic conditions.

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.