Problem 36 - Chemistry and Chemical Reactivity

Question

A self-contained underwater breathing apparatus uses canisters containing potassium superoxide. The superoxide consumes the \(\mathrm{CO}_{2}\) exhaled by a person and replaces it with oxygen. $$ 4 \mathrm{KO}_{2}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) $$ What mass of \(\mathrm{KO}_{2}\), in grams, is required to react with \(8.90 \mathrm{L}\) of \(\mathrm{CO}_{2}\) at \(22.0^{\circ} \mathrm{C}\) and \(767 \mathrm{mm}\) Hg?

Step-by-Step Solution

Answer

52.2 g of KO2 is needed.

Step 1: Understand the Chemical Reaction

The chemical reactor uses potassium superoxide (\( ext{KO}_2\)) to consume carbon dioxide (\( ext{CO}_2\)) and release oxygen (\( ext{O}_2\)). The balanced equation is: \[4 \mathrm{KO}_2(\mathrm{s}) + 2 \mathrm{CO}_2(\mathrm{g}) \rightarrow 2 \mathrm{K}_2 \mathrm{CO}_3(\mathrm{s}) + 3 \mathrm{O}_2(\mathrm{g})\]. The reaction shows that 4 moles of \( ext{KO}_2\) react with 2 moles of \( ext{CO}_2\).

Step 2: Convert the Volume of CO2 to Moles

First, use the ideal gas law \(PV = nRT\) to find the moles of carbon dioxide. Convert the temperature from Celsius to Kelvin: \(T = 22 + 273.15 = 295.15\,\mathrm{K}\). Convert the pressure from mm Hg to atm: \(767\,\mathrm{mm}\,\mathrm{Hg} \times \frac{1\,\mathrm{atm}}{760\,\mathrm{mm}\,\mathrm{Hg}} = 1.00921\,\mathrm{atm}\). Then solve for \(n\):\[n = \frac{PV}{RT} = \frac{(1.00921\,\mathrm{atm}) \times (8.90\,\mathrm{L})}{(0.0821\,\mathrm{L}\cdot\mathrm{atm}/\mathrm{K}\cdot\mathrm{mol})(295.15\,\mathrm{K})}= 0.367\,\mathrm{mol}\,\mathrm{CO}_2\].

Step 3: Use Stoichiometry to Find Moles of KO2

According to the balanced equation, 4 moles of \( ext{KO}_2\) react with 2 moles of \( ext{CO}_2\). Calculate the moles of \( ext{KO}_2\) needed:\[\text{Moles of KO}_{2} = 0.367 \, \text{mol CO}_{2} \times \frac{4\,\text{mol KO}_{2}}{2\,\text{mol CO}_{2}} = 0.734 \, \text{mol KO}_{2}\].

Step 4: Convert Moles of KO2 to Mass

Find the molar mass of \( ext{KO}_2\): \( \text{K: } 39.10\,\text{g/mol}, \text{O: } 16.00\,\text{g/mol}\). \( ext{Molar mass of KO}_2 = 39.10 + 2 \times 16.00 = 71.10\,\text{g/mol}\). Convert moles of \(\text{KO}_2\) to grams using its molar mass:\[\text{Mass of KO}_{2} = 0.734\,\text{mol KO}_{2} \times 71.10\,\text{g/mol KO}_{2} = 52.2\,\text{g}\].

Step 5: Interpret the Result

The mass of \(\text{KO}_2\) required to react completely with \(8.90\,\text{L}\) of \(\text{CO}_2\) under the given conditions is \(52.2\,\mathrm{g}\). This number represents the amount of potassium superoxide needed to maintain the breathing cycle within the apparatus.

Key Concepts

Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It helps you understand the proportions of reactants and products. In the reaction between potassium superoxide (\(\text{KO}_2\)) and carbon dioxide (\(\text{CO}_2\)), stoichiometry tells us how much of each substance is involved. The balanced equation shows that 4 moles of \(\text{KO}_2\) react with 2 moles of \(\text{CO}_2\). This relationship is crucial for determining how much \(\text{KO}_2\) you need when you have a certain amount of \(\text{CO}_2\).

• If you know the amount of one reactant, you can find the amount of other reactants needed using these mole ratios.
• Stoichiometry uses these ratios to ensure a reaction is feasible and efficient.

By using stoichiometry, you can calculate the mass, moles, and volume of substances involved in chemical reactions.

Chemical Reaction Balancing

Chemical reaction balancing is adjusting the coefficients in a chemical equation so that the number of each type of atom is the same on both sides. This is important because matter cannot be created or destroyed. In our reaction:\[4 \text{KO}_2(\text{s}) + 2 \text{CO}_2(\text{g}) \rightarrow 2 \text{K}_2 \text{CO}_3(\text{s}) + 3 \text{O}_2(\text{g})\]
The equation is balanced because:

• There are 4 potassium (K), 8 oxygen (O), and 2 carbon (C) atoms on each side.
• Balancing ensures that the chemical reaction preserves mass and energy.

This process lets us predict the amounts of substances consumed or produced in a reaction precisely.

Gas Laws

Gas laws describe how gases behave and help us calculate the volume, temperature, pressure, and amount of gas involved. For this problem, the **Ideal Gas Law** is key: \(PV = nRT\). This formula allows us to connect pressure (P), volume (V), and temperature (T) to find the number of moles (n).
Here's how it works:

• Convert all measurements to proper units; temperature in Kelvin and pressure in atm.
• Use R, the ideal gas constant, which is typically 0.0821 \(\text{L} \cdot \text{atm}/\text{K} \cdot \text{mol}\).

In this situation, the formula is rearranged to solve for moles of \(\text{CO}_2\), giving you insight into how much of the gas is involved under certain conditions.

Molar Mass Calculation

Molar mass is the weight of one mole of a chemical substance. It's essential for converting between moles and grams. In the example with \(\text{KO}_2\), we find the molar mass by adding the atomic masses of potassium and oxygen:

• Potassium (K) has an atomic mass of 39.10 g/mol.
• Each oxygen (O) has an atomic mass of 16.00 g/mol.

The formula for \(\text{KO}_2\) adds up to a molar mass of 71.10 g/mol (39.10 + 2×16.00).
Knowing the molar mass, we can convert from moles to grams, which helps find out how much \(\text{KO}_2\) is required for the reaction.