Question

Some older emergency oxygen masks containing potassium superoxide, \(\mathrm{KO}_{2}\) which reacts with \(\mathrm{CO}_{2}\) and water in exhaled air to produce oxygen according to the given equation. If a person exhales \(0.667 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) per minute, how many grams of \(\mathrm{KO}_{2}\) are consumed in \(5.0\) minutes? $$ 4 \mathrm{KO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+4 \mathrm{CO}_{2} \longrightarrow 4 \mathrm{KHCO}_{3}+3 \mathrm{O}_{2} $$ (a) \(10.7\) (b) \(0.0757\) (c) \(1.07\) (d) \(5.38\)

Short answer

5.38 grams of KO2 are consumed in 5 minutes.

Step-by-step solution

Determine the total mass of CO2 exhaled

The problem states that a person exhales 0.667 g of CO2 per minute. To find the total mass of CO2 exhaled in 5 minutes, multiply that amount by 5: Total mass of CO2 = 0.667 g/min * 5 min = 3.335 g.

Use the stoichiometry of the reaction

The reaction equation given is: \(4 \mathrm{KO}_2 + 2 \mathrm{H}_2\mathrm{O} + 4 \mathrm{CO}_2 \longrightarrow 4 \mathrm{KHCO}_3 + 3 \mathrm{O}_2\). According to the balanced chemical equation, 4 moles of CO2 react with 4 moles of KO2. We use this ratio to calculate the moles of KO2 needed for the reaction.

Calculate the moles of CO2 exhaled

First, calculate the moles of CO2 using its molar mass (44.01 g/mol): Moles of CO2 = Total mass of CO2 / Molar mass of CO2 = 3.335 g / 44.01 g/mol = 0.0757 moles.

Calculate the moles of KO2 needed

Using the stoichiometric ratio, 1 mole of CO2 reacts with 1 mole of KO2, so the moles of KO2 needed is equal to the moles of CO2 exhaled: Moles of KO2 needed = 0.0757 moles.

Calculate the mass of KO2 needed

Finally, calculate the mass of KO2 using its molar mass (71.10 g/mol): Mass of KO2 = Moles of KO2 * Molar mass of KO2 = 0.0757 moles * 71.10 g/mol = 5.38 g.

Key concepts

Chemical Reaction Equations

Understanding chemical reaction equations is essential for solving stoichiometry problems.
A chemical equation represents the transformation of reactants into products, indicating the exact number of molecules or moles involved. In the given exercise, the reaction equation is:
\[4 \mathrm{KO}_2 + 2 \mathrm{H}_2\mathrm{O} + 4 \mathrm{CO}_2 \longrightarrow 4 \mathrm{KHCO}_3 + 3 \mathrm{O}_2\]
This equation shows that 4 moles of potassium superoxide (\(\mathrm{KO}_2\)) react with 2 moles of water (\(\mathrm{H}_2\mathrm{O}\)) and 4 moles of carbon dioxide (\(\mathrm{CO}_2\)) to produce 4 moles of potassium hydrogen carbonate (\(\mathrm{KHCO}_3\)) and 3 moles of oxygen gas (\(\mathrm{O}_2\)). Understanding how to interpret and balance chemical equations is a foundational skill for any chemistry student.

Importance of a Balanced Equation

A balanced reaction equation respects the law of conservation of mass, which states that the mass of reactants must equal the mass of products. Balancing is done by adjusting coefficients (the numbers in front of molecules) to make sure the number of atoms for each element is equal on both sides of the equation.

Molar Mass Calculations

Molar mass calculations are a vital part of chemical stoichiometry, allowing us to convert between mass and moles of a substance.
The molar mass is the weight of one mole of a chemical compound or element and is usually expressed in grams per mole (g/mol). For instance, in the provided exercise, the molar masses of carbon dioxide (\(\mathrm{CO}_2\)) and potassium superoxide (\(\mathrm{KO}_2\)) are given as 44.01 g/mol and 71.10 g/mol, respectively. To calculate the moles of \(\mathrm{CO}_2\) exhaled based on the mass, we use the formula:
\[\text{Moles of CO}_2 = \frac{\text{Total mass of CO}_2}{\text{Molar mass of CO}_2}\]
Knowing the molar mass is crucial for converting the mass of reactants to moles, which then allows for the use of stoichiometric ratios in reactions to determine the mass of other substances involved.

Stoichiometric Calculations

Stoichiometric calculations involve using the relationships between the reactants and products in a balanced chemical equation to calculate unknown quantities.
In stoichiometry, the mole ratio derived from the balanced equation is the key to solving problems. These ratios tell us how many moles of one substance will react or form from a given number of moles of another substance. In the oxygen mask problem, the stoichiometric ratio of \(\mathrm{CO}_2\) to \(\mathrm{KO}_2\) is 1:1. Therefore, the moles of \(\mathrm{KO}_2\) needed for the reaction are equal to the moles of \(\mathrm{CO}_2\) exhaled. This step is where the mole concept unites with chemical reactions, bridging the gap between the microscopic world of atoms and molecules and the macroscopic world that we can measure.

Applying Stoichiometry to Find Mass

After calculating the moles of \(\mathrm{KO}_2\) needed based on the moles of exhaled \(\mathrm{CO}_2\), we then convert the moles back to mass using the molar mass of \(\mathrm{KO}_2\):
\[\text{Mass of KO}_2 = \text{Moles of KO}_2 \times \text{Molar mass of KO}_2\]
Such stoichiometric calculations are fundamental in predicting product formation and reactant consumption, which are key in many practical applications such as medical oxygen generation systems mentioned in the exercise.