Question

A solution of potassium permanganate reacts with oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) to form carbon dioxide and solid manganese(IV) oxide \(\left(\mathrm{MnO}_{2}\right)\) according to the equation $$\begin{aligned} 3 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+2 \mathrm{MnO}_{4}^{-}(a q) &+2 \mathrm{H}^{+}(a q) \longrightarrow \\ & 6 \mathrm{CO}_{2}(g)+2 \mathrm{MnO}_{2}(s)+4 \mathrm{H}_{2} \mathrm{O} \end{aligned}$$ (a) If \(20.0 \mathrm{~mL}\) of \(0.300 \mathrm{M}\) potassium permanganate are required to react with \(13.7 \mathrm{~mL}\) of oxalic acid, what is the molarity of the oxalic acid? (b) What is the mass of manganese(IV) oxide formed?

Short answer

Answer: (a) The molarity of the oxalic acid is 0.656 M. (b) The mass of manganese(IV) oxide formed is 0.521 g.

Step-by-step solution

Calculate moles of potassium permanganate

We can use the given volume and molarity of potassium permanganate to determine the moles of potassium permanganate present by using the formula: moles = molarity × volume (convert the volume to L). Moles of potassium permanganate = \((0.300 \thinspace M)(0.0200 \thinspace L)\) Moles of potassium permanganate = \(0.00600 \thinspace mol\)

Use stoichiometry to find moles of oxalic acid required

Using the balanced chemical equation, we can see that it takes 3 moles of oxalic acid to react with 2 moles of potassium permanganate. Thus, we can calculate the moles of oxalic acid required: \(\frac{3 \thinspace mol \thinspace H_2C_2O_4}{2 \thinspace mol \thinspace MnO_4^-} = \frac{x \thinspace mol \thinspace H_2C_2O_4}{0.00600 \thinspace mol \thinspace MnO_4^-}\) Solving for x, which represents the moles of oxalic acid: \(x = \frac{3 \thinspace mol \thinspace H_2C_2O_4 \times 0.00600 \thinspace mol \thinspace MnO_4^-}{2 \thinspace mol \thinspace MnO_4^-} = 0.00900 \thinspace mol \thinspace H_2C_2O_4\)

Find the molarity of the oxalic acid

Now, we can use the moles of oxalic acid and the given volume of oxalic acid to calculate its molarity using the formula: molarity = moles/volume (convert the volume to L). Molarity of oxalic acid = \(\frac{0.00900 \thinspace mol}{0.0137 \thinspace L}\) Molarity of oxalic acid = \(0.656 \thinspace M\)

Calculate moles of manganese(IV) oxide formed

Using the balanced chemical equation, we can find out how many moles of manganese(IV) oxide are formed by reacting 0.00600 moles of potassium permanganate. \(\frac{2 \thinspace mol \thinspace MnO_2}{2 \thinspace mol \thinspace MnO_4^-} = \frac{y \thinspace mol \thinspace MnO_2}{0.00600 \thinspace mol \thinspace MnO_4^-}\) Solving for y, which represents the moles of manganese(IV) oxide: \(y = \frac{2 \thinspace mol \thinspace MnO_2 \times 0.00600 \thinspace mol \thinspace MnO_4^-}{2 \thinspace mol \thinspace MnO_4^-} = 0.00600 \thinspace mol \thinspace MnO_2\)

Calculate the mass of manganese(IV) oxide formed

Now that we have the moles of manganese(IV) oxide formed, we can find the mass by using the molar mass of manganese(IV) oxide (MnO2), which is 86.9368 g/mol: Mass of manganese(IV) oxide = \((0.00600 \thinspace mol) (86.9368 \thinspace g/mol)\) Mass of manganese(IV) oxide = \(0.521 \thinspace g\) To summarize: (a) The molarity of the oxalic acid is \(0.656 \thinspace M\). (b) The mass of manganese(IV) oxide formed is \(0.521 \thinspace g\).

Key concepts

Molarity Calculation

When studying chemistry, understanding the concentration of a solution is crucial. Molarity, which is often denoted as M, is one such measure of concentration that refers to the number of moles of a solute per liter of solution. It's calculated using the formula:

\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]
Suppose we have a solution of potassium permanganate (\text{KMnO}_{4}) with a molarity of 0.300 M and a volume of 20.0 mL. To find the number of moles of \text{KMnO}_{4}, we convert the volume from milliliters to liters (20.0 mL = 0.0200 L) and then multiply by the molarity:

\[\text{Moles of KMnO}_{4} = 0.300 \, M \times 0.0200 \, L = 0.00600 \, mol\]
This molarity calculation is pivotal as it's used to determine stoichiometry in chemical reactions. When the molarity of oxalic acid is sought, similarly, its calculated moles and known volume in liters allow the determination of its concentration in the solution.

Balanced Chemical Equation

The reaction between potassium permanganate and oxalic acid to produce carbon dioxide, manganese(IV) oxide, and water can be described with a balanced chemical equation. A balanced chemical equation ensures that the number of atoms of each element is the same on both the reactants' and products' sides, thereby satisfying the law of conservation of mass. In our chemical reaction:

\[3 \text{H}_{2}\text{C}_{2}\text{O}_{4} (aq) + 2 \text{MnO}_{4}^{-}(aq) + 2 H^{+}(aq) \rightarrow 6 \text{CO}_{2}(g) + 2 \text{MnO}_{2}(s) + 4 \text{H}_{2}\text{O}(l)\]
Every coefficient in this balanced equation represents a mole ratio, which is essential for stoichiometric calculations. For example, the coefficient '3' before \text{H}_{2}\text{C}_{2}\text{O}_{4} indicates that three moles of oxalic acid are needed for every two moles of potassium permanganate. By keeping this equation balanced, it provides a clear map of how substances interact and transform in chemical reactions.

Mole-to-Mole Ratio

The mole-to-mole ratio is derived from a balanced chemical equation and is essential for stoichiometry, offering a quantitative understanding of how reactants and products interrelate in a chemical reaction. For instance, in our problem, the balanced equation indicates a mole ratio of 3 moles of oxalic acid (\text{H}_{2}\text{C}_{2}\text{O}_{4}) to 2 moles of potassium permanganate (\text{MnO}_{4}^{-}). This ratio allows us to calculate the amount of one chemical in moles when we know the amount of another chemical in the reaction in moles.

For example, to find out how many moles of oxalic acid are needed to react with 0.00600 moles of potassium permanganate, we set up a proportion:

\[\frac{3 \text{ moles } \text{H}_{2}\text{C}_{2}\text{O}_{4}}{2 \text{ moles } \text{MnO}_{4}^{-}} = \frac{\text{x moles } \text{H}_{2}\text{C}_{2}\text{O}_{4}}{0.00600 \text{ moles } \text{MnO}_{4}^{-}}\]
Solving this ratio for x gives us the moles of oxalic acid required for the reaction. Similarly, we can use the mole-to-mole ratio to determine the quantity of manganese(IV) oxide formed in a reaction. Mastering mole-to-mole conversions is fundamental in predicting the outcomes of chemical reactions and in manufacturing chemicals with precision.