Problem 64 - General Chemistry: Principles and Modern Applications

Question

A 0.589 g sample of pyrolusite ore (impure \(\mathrm{MnO}_{2}\) ) is treated with \(1.651 \mathrm{g}\) of oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)\) in an acidic medium (reaction 1). Following this, the excess oxalic acid is titrated with \(30.06 \mathrm{mL}\) of \(0.1000 \mathrm{M}\) \(\mathrm{KMnO}_{4}\) (reaction 2). What is the mass percent of \(\mathrm{MnO}_{2}\) in the pyrolusite? The following equations are neither complete nor balanced. (1) \(\quad \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{s}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})\) (2) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{MnO}_{4}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})\)

Step-by-Step Solution

Answer

The mass percent of MnO2 in the pyrolusite is 41.22%.

Step 1: Balance the reaction equations

Reaction 1: 2H2C2O4(aq) + MnO2(s) -> Mn2+(aq) + CO2(g) + 2H2O(l) + 2e-. This reaction involves reduction. Reaction 2: 5H2C2O4(aq) + 2MnO4-(aq) -> 10CO2(g) + 8H2O(l) + 2Mn2+(aq). This reaction involves oxidation.

Step 2: Calculate the moles of KMnO4

KMnO4 reacts with the excess oxalic acid. Given the molarity (0.1000 M), and volume (30.06 mL or 0.03006 L), the moles can be determined: mol KMnO4 = Molarity * Volume = 0.1000 M * 0.03006 L = 0.003006 mol KMnO4.

Step 3: Determine the moles of H2C2O4 reacted with KMnO4

From the balanced equation 5H2C2O4 -> 2 KMnO4, every 5 moles of oxalic acid react with 2 moles of KMnO4. Therefore, the moles of oxalic acid that reacted with KMnO4 are: mol H2C2O4 = (mol KMnO4 * 5) / 2 = 0.003006 * 5 / 2 = 0.007515 mol H2C2O4.

Step 4: Calculate the total moles of H2C2O4

The total moles of oxalic acid are given by the mass (in g) divided by the molar mass. Given mass is 1.651g and molar mass is 126.07 g/mol, therefore the moles = mass / molar mass = 1.651g / 126.07g/mol = 0.0131 mol H2C2O4.

Step 5: Determine the moles of H2C2O4 reacted with MnO2

The moles of oxalic acid reacted with MnO2 are given by the difference between total moles of oxalic acid and those reacted with KMnO4: mol H2C2O4 = total mol H2C2O4 - mol reacted with KMnO4 = 0.0131 - 0.007515 = 0.005585 mol H2C2O4.

Step 6: Calculate the moles and mass of MnO2

From the balanced equation, every 2 moles of oxalic acid react with 1 mole of MnO2. Therefore, the moles of MnO2 that reacted are given by: mol MnO2 = mol H2C2O4 / 2 = 0.005585 / 2 = 0.0027925 mol MnO2. The mass of MnO2 is given by multiply this with its molar mass: mass = mol MnO2 * Molar mass MnO2 = 0.0027925 mol * 86.94 g/mol = 0.2428 g.

Step 7: Calculate the mass percent of MnO2

The mass percent of MnO2 is given by the mass of MnO2 reacted divided by the total mass of the sample multiplied by 100%: MnO2 mass percent = (mass MnO2/ sample mass) * 100% = (0.2428 g / 0.589 g) * 100% = 41.22%

Key Concepts

Redox Reactions

Redox reactions, also known as oxidation-reduction reactions, play a crucial role in the exercise of determining the mass percent of \(\mathrm{MnO}_2\) in a pyrolusite ore sample. These reactions involve the transfer of electrons between chemical species. In the problem, we have two reactions: one is a reduction process where oxalic acid (\(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4\)) reacts with pyrolusite (\(\mathrm{MnO}_2\)), and the other is an oxidation process involving oxalic acid and potassium permanganate (\(\mathrm{KMnO}_4\)).

The reduction reaction sees \(\mathrm{MnO}_2\) gaining electrons, forming Mn²⁺ ions, while oxalic acid is oxidized to carbon dioxide. Conversely, in the oxidation reaction, oxalic acid loses electrons and \(\mathrm{MnO}_4^-\) is reduced.
These balanced reactions allow us to understand the stoichiometry needed to determine the amount of \(\mathrm{MnO}_2\) present. Grasping the fundamental concept of electron transfer in redox reactions is essential for solving mass percent calculations effectively.

Titration

Titration is a process used to determine the concentration of a solute in a solution. In this exercise, titration is employed to find the amount of excess oxalic acid that remains after reacting with \(\mathrm{MnO}_2\).

Potassium permanganate (\(\mathrm{KMnO}_4\)) is the titrant, and it is added to the solution containing excess oxalic acid until a reaction endpoint is reached, indicated by a color change.
The volume of \(\mathrm{KMnO}_4\) used (30.06 mL) combined with its concentration (0.1000 M) allows us to calculate the moles of \(\mathrm{KMnO}_4\), which further helps in establishing how much oxalic acid reacted with \(\mathrm{MnO}_4^-\). This back titration approach simplifies determining the oxalic acid that reacted with \(\mathrm{MnO}_2\) initially, thereby aiding in mass calculations.

Moles and Molarity

Understanding moles and molarity is fundamental in chemistry, especially in this exercise. Moles provide a way to count particles, which is crucial for stoichiometric calculations. Molarity, defined as moles of solute per liter of solution, enables us to calculate and relate the quantity of reactants and products.

For example, determining the moles of \(\mathrm{KMnO}_4\) using its concentration (0.1000 M) and volume (30.06 mL) is needed to deduce how much oxalic acid was reacted.
Also, knowing the total moles of oxalic acid initially present, calculated by dividing its mass by molar mass, allows us to find how many moles participated in redox reactions. This together enables us to accurately compute the amount of \(\mathrm{MnO}_2\) in the ore sample.
Keep this in mind: Moles connect the macroscopic measurements to the atomic scale, making calculations about reactions possible and accurate.

Stoichiometry

Stoichiometry is the part of chemistry that involves calculating the relationships of reactants and products in chemical reactions. It is the backbone for our solution in finding out how much \(\mathrm{MnO}_2\) was present in the sample.

We employ stoichiometry to analyze the balanced equations of the redox reactions. By understanding the mole ratios from the balanced equations, like the 2:1 ratio between \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4\) and \(\mathrm{MnO}_2\) in one reaction, we can calculate the moles of \(\mathrm{MnO}_2\) reacted with oxalic acid.
This information is crucial in finally calculating the mass and subsequently the mass percent of \(\mathrm{MnO}_2\) in the ore sample.
Notably, stoichiometry transforms theoretical chemistry into practical applications, letting us determine quantities and make predictions based on substance amounts.