Question

What mass of \(\mathrm{KO}_{2}\) is required to remove \(90.0 \%\) of the \(\mathrm{CO}_{2}\) from a sample of \(1.00 \mathrm{~L}\) of exhaled air \(\left(37^{\circ} \mathrm{C}, 1.00 \mathrm{~atm}\right)\) containing \(5.00\) mole percent \(\mathrm{CO}_{2}\) ?

Short answer

To remove 90% of CO₂ from a 1.00 L sample of exhaled air containing 5.00 mole percent CO₂, approximately 0.2529 g of KO₂ is required.

Step-by-step solution

Find moles of CO₂ in exhaled air

Using the ideal gas law, we can calculate the number of moles of air: \(PV = nRT\) Given that pressure P = \(1.00\, \mathrm{atm}\), volume V = \(1.00\, \mathrm{L}\), temperature T = \(37^\circ \mathrm{C}\) (convert to Kelvin: \(273 + 37 = 310 \, \mathrm{K}\)) and the gas constant R = \(0.0821\, \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}}\). \(n = \frac{PV}{RT} = \frac{1.00\,\mathrm{atm} \cdot 1.00\,\mathrm{L}}{0.0821\, \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}} \cdot 310\,\mathrm{K}} = 0.0395 \, \mathrm{mol}\) Now we find the moles of CO₂ by using the mole percent information: Moles of \(\mathrm{CO}_2\) = \((0.0395\, \mathrm{mol})\left(\frac{5.00}{100}\right) = 0.001975\, \mathrm{mol}\)

Calculate moles of CO₂ to be removed

We are given that \(90\%\) of \(\mathrm{CO}_2\) must be removed. So, we calculate the moles of \(\mathrm{CO}_2\) that needs to be removed: Moles of \(\mathrm{CO}_2\) to be removed = \(0.001975\, \mathrm{mol} \cdot 0.90 = 0.0017775\, \mathrm{mol}\)

Find the stoichiometry of the reaction involving KO₂ and CO₂

The reaction between \(\mathrm{KO}_2\) and \(\mathrm{CO}_2\) can be represented as: \(2\, \mathrm{KO}_2 + \mathrm{CO}_2 \to \mathrm{K_2CO}_3 + 1.5\, \mathrm{O}_2\) From the stoichiometry of the reaction, we see that 2 moles of \(\mathrm{KO}_2\) react with 1 mole of \(\mathrm{CO}_2\).

Calculate the moles of KO₂ required

Based on the stoichiometry of the reaction, we can calculate the moles of \(\mathrm{KO}_2\) required to remove \(0.0017775\, \mathrm{mol}\) of \(\mathrm{CO}_2\): Moles of \(\mathrm{KO}_2\) required = \(\frac{2\, \mathrm{mol} \, \mathrm{KO}_2}{1\, \mathrm{mol} \, \mathrm{CO}_2} \cdot 0.0017775\, \mathrm{mol} \, \mathrm{CO}_2\) = \(0.003555\, \mathrm{mol}\, \mathrm{KO}_2\)

Calculate the mass of KO₂

Now that we know the moles of \(\mathrm{KO}_2\), we can calculate the mass by using the molar mass of \(\mathrm{KO}_2\): Molar mass of \(\mathrm{KO}_2 = 39.10(\mathrm{K}) + 15.999(\mathrm{O}_2) = 71.10\, \mathrm{g/mol}\). Mass of \(\mathrm{KO}_2\) required = \((0.003555\, \mathrm{mol}\, \mathrm{KO}_2)\left(\frac{71.10\, \mathrm{g}}{1\, \mathrm{mol}\, \mathrm{KO}_2}\right) = 0.2529\,\mathrm{g}\) So, the mass of \(\mathrm{KO}_2\) required to remove \(90.0\%\) of the \(\mathrm{CO}_2\) from a sample of \(1.00\, \mathrm{L}\) of exhaled air containing \(5.00\) mole percent \(\mathrm{CO}_2\) is approximately \(0.2529\, \mathrm{g}\).

Key concepts

Molar Mass Calculation

Molar mass calculation is a fundamental concept in chemistry that allows us to convert between the mass of a substance and the number of moles. The molar mass is the mass of one mole of a substance (in grams per mole, g/mol), which numerically is equivalent to the substance's atomic or molecular mass.

To calculate the molar mass, we sum the atomic masses of all the atoms present in a molecule. For example, for potassium superoxide (\textbf{KO}\(_2\)), the molar mass is calculated by adding the atomic mass of potassium (K, approximately 39.10 g/mol) to the mass of two oxygen atoms (O, approximately 16.00 g/mol each), giving us a total of approximately 71.10 g/mol. This is vital for stoichiometric calculations, as it allows us to find out how much of a substance is needed or produced in a chemical reaction.

Ideal Gas Law

The ideal gas law, represented by the equation PV = nRT, describes the relationship between pressure (P), volume (V), number of moles (n), temperature (T), and the ideal gas constant (R), which is 0.0821 atm L/mol K.

When dealing with gases, this law allows us to determine the amount of gas (in moles) if we know the temperature, volume, and pressure of the gas in question. For instance, in the exercise, we used the ideal gas law to calculate the number of moles of exhaled air at 37°C and 1 atm pressure. Temperature is always considered in Kelvin in these calculations, which means we need to convert Celsius to Kelvin by adding 273.15. This law is especially critical when analyzing gases involved in chemical reactions, such as determining the quantity of \textbf{CO}\(_2\) in exhaled air, which forms the basis for further stoichiometric calculations.

Chemical Reaction Equation

A chemical reaction equation gives a symbolic representation of a chemical reaction. It shows the reactants that are consumed and the products that are formed, along with their stoichiometric coefficients, which tell us the proportions in which the substances react and are produced.

In the exercise, the chemical reaction equation between potassium superoxide (\textbf{KO}\(_2\)) and carbon dioxide (\textbf{CO}\(_2\)) is used to determine the amount of \textbf{KO}\(_2\) required to remove a certain percentage of \textbf{CO}\(_2\):

2 \textbf{KO}\(_2\) + \textbf{CO}\(_2\) \rightarrow \textbf{K}\(_2\)\textbf{CO}\(_3\) + 1.5 \textbf{O}\(_2\)

. This ratio (2 moles of \textbf{KO}\(_2\) for every mole of \textbf{CO}\(_2\)) is crucial to solve the problem, as it lets us calculate the exact amount of \textbf{KO}\(_2\) needed to achieve the desired reaction.