Problem 6 - Chemical Principles: The Quest for Insight

Question

Potassium superoxide, \(\mathrm{KO}_{2}\), is utilized in closedsystem breathing apparatus to remove carbon dioxide and water from exhaled air. The removal of water generates oxygen for breathing by the reaction $$ 4 \mathrm{KO}_{2}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 3 \mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{KOH}(\mathrm{s}) $$ The potassium hydroxide removes carbon dioxide from the apparatus by the reaction $$ \mathrm{KOH}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{KHCO}_{3}(\mathrm{~s}) $$ (a) What mass of potassium superoxide generates \(20.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ? (b) What mass of \(\mathrm{CO}_{2}\) can be removed from the apparatus by \(100 \mathrm{~g}\) of \(\mathrm{KO}_{2}\) ?

Step-by-Step Solution

Answer

a) 59.27 g of KO2 is required to generate 20.0 g of O2. b) 61.89 g of CO2 can be removed by 100 g of KO2.

Step 1: Calculate the Molar Mass of the Substances Involved

Determine the molar mass of the reactant and product involved in the first reaction (4 KO2 + 2 H2O -> 3 O2 + 4 KOH). The molar mass of KO2 (potassium superoxide) and O2 (oxygen) is needed for this calculation. Use the periodic table to find the atomic masses for potassium (K), oxygen (O), and hydrogen (H). Potassium superoxide (KO2) has a molar mass of (K = 39.10 g/mol) + (2 x O = 2 x 16.00 g/mol) = 71.10 g/mol. Dioxygen (O2) has a molar mass of 2 x 16.00 g/mol = 32.00 g/mol.

Step 2: Calculate the Mass of Potassium Superoxide Required to Generate 20.0g of O2

Use the stoichiometry of the reaction to relate the mass of O2 to the mass of KO2. From the balanced chemical equation, 4 moles of KO2 produce 3 moles of O2. Therefore, using the molar mass of O2, you can calculate the moles of O2 produced from 20.0 g of O2: moles O2 = mass O2 / molar mass O2 = 20.0 g / 32.00 g/mol = 0.625 mol. Now, use the stoichiometry of the balanced equation to find the moles of KO2 required to produce 0.625 moles of O2: (4 moles KO2 / 3 moles O2) x 0.625 mol O2 = 0.833 mol KO2. Then, convert the moles of KO2 to grams: mass KO2 = moles KO2 x molar mass KO2 = 0.833 mol x 71.10 g/mol = 59.27 g.

Step 3: Calculate the Mass of CO2 That Can Be Removed by 100g of KO2

First, find the number of moles of KO2 in 100 g: moles of KO2 = 100 g / 71.10 g/mol = 1.406 mol. Then, use the stoichiometry from the second reaction (KOH + CO2 -> KHCO3) to find the moles of CO2 that can be reacted: because 1 mole of KOH reacts with 1 mole of CO2, and we know the reaction with KO2 produces KOH in a 1:1 mole ratio, the moles of KO2 will be equal to the moles of CO2. Thus, 1.406 mol of KO2 will react with 1.406 mol of CO2. Finally, calculate the mass of CO2 that can be removed: mass CO2 = moles CO2 x molar mass CO2. The molar mass of CO2 (C = 12.01 g/mol, O2 = 2 x 16.00 g/mol) is 44.01 g/mol. Therefore, mass of CO2 = 1.406 mol x 44.01 g/mol = 61.89 g.

Key Concepts

Chemical Reaction Balancing

Understanding how to balance chemical reactions is fundamental to the study of chemistry. When we balance equations, we're ensuring that the number of atoms for each element is the same on both sides of the equation. This principle is based on the law of conservation of mass, which states that in a closed system, mass cannot be created or destroyed.

In our exercise example, the first reaction depicts potassium superoxide reacting with water to produce oxygen and potassium hydroxide. To achieve an equal number of atoms for each element on both sides of the equation, we balance the reaction by adding coefficients, which are the numbers placed before the substances. The properly balanced equation will determine the stoichiometric relationships, which are crucial for the subsequent calculations.

For instance, balancing the oxygen atoms can be tricky since they appear in all reactants and products. The balancing act ensures that we have equal numbers of potassium, oxygen, and hydrogen atoms before and after the reaction. This balanced equation allows us to relate the amounts of reactants to products quantitatively.

Molar Mass Calculation

The molar mass of a compound is the sum of the total mass in grams of all the atoms in a mole of that substance. It's a crucial conversion factor that bridges the atomic world to the one we can measure in the lab. For molar mass calculations, we rely on the periodic table for the atomic masses of each element.

In the original problem, calculating the molar mass of potassium superoxide (\textbf{KO2}) involves adding the atomic mass of one potassium atom to twice the mass of an oxygen atom, presented as: \( \text{Molar mass of KO2} = (39.10 + 2 \times 16.00) \text{ g/mol} = 71.10 \text{ g/mol} \).

Understanding molar mass is essential for converting between the mass of a substance and the number of moles, which then allows us to use the balanced chemical equation for stoichiometric calculations. This concept is continually applied in stoichiometry to predict the amounts of substances consumed and produced in a given chemical reaction.

Mole-to-Mass Conversion

Mole-to-mass conversion is a process that allows chemists to convert the number of moles of a substance to its corresponding mass, using the molar mass as a conversion factor. This conversion is vital in quantifying the amount of reagents and products in a chemical reaction.

In the original problem, once we've determined the number of moles of oxygen produced, we use the stoichiometry of the balanced equation to find out the number of moles of potassium superoxide needed. The next step is converting these moles back to grams to find the mass required for the reaction, using the formula: \( \text{mass} = \text{moles} \times \text{molar mass} \).

For example, to generate 20.0 g of oxygen, we calculated the number of moles of \textbf{KO2} required and then converted those moles into grams, answering part (a) of the exercise. This process of mole-to-mass conversion provides the practical mass that can be measured and used by scientists in experiments and industrial applications.