Problem 100 - General Chemistry: Principles and Modern Applications

Question

In the reaction of \(\mathrm{CO}_{2}(\mathrm{g})\) and solid sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right),\) solid sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) and oxy- gen gas are formed. This reaction is used in submarines and space vehicles to remove expired \(\mathrm{CO}_{2}(\mathrm{g})\) and to generate some of the \(\mathrm{O}_{2}(\mathrm{g})\) required for breathing. Assume that the volume of gases exchanged in the lungs equals \(4.0 \mathrm{L} / \mathrm{min},\) the \(\mathrm{CO}_{2}\) content of expired air is \(3.8 \% \mathrm{CO}_{2}\) by volume, and the gases are at \(25^{\circ} \mathrm{C}\) and \(735 \mathrm{mmHg}\). If the \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) in the above reaction are measured at the same temperature and pressure, (a) how many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) are produced per minute and \((\mathrm{b})\) at what rate is the \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s})\) consumed, in grams per hour?

Step-by-Step Solution

Answer

a) The volume of \(\mathrm{O}_{2}\) produced per minute is approximately 76 mL. b) The rate at which the \(\mathrm{Na}_{2} \mathrm{O}_{2}\) is consumed is approximately 1.165 g per hour.

Step 1: Writing the balanced chemical equation

The chemical equation for this reaction is \[\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s}) + 2\mathrm{CO}_{2}(\mathrm{g}) \rightarrow 2\mathrm{Na}_{2} \mathrm{CO}_{3}\left(\mathrm{s}\right) + \mathrm{O}_{2}(\mathrm{g})\] Remember to balance the equation, each side of the equation must have the same number of atoms of each element.

Step 2: Calculate the amount of CO2 exhaled per minute

The volume of gases exchanged in the lungs equals \(4.0 \mathrm{L} / \mathrm{min}\) and \(3.8 \% \) of this volume is CO2. So, the volume of CO2 exhaled per minute can be calculated as \(4.0 \mathrm{L/min} \times 0.038 = 0.152 \mathrm{L/min} = 152 \mathrm{mL/min}\).

Step 3: Calculate the volume of O2 produced per minute

From the balanced chemical equation, it can be seen that 1 mole of O2 is produced from 2 moles of CO2. Therefore, the volume of O2 produced per minute is half the volume of CO2 exhaled per minute, which is \(152 / 2 = 76 \mathrm{mL/min}\).

Step 4: Calculate the rate of Na2O2 consumed

The balanced equation also shows that 1 mole of Na2O2 reacts with 2 moles of CO2. Therefore, the moles of Na2O2 consumed per minute is half the moles of CO2 exhaled per minute. Using the molar volume of a gas at the given conditions (approximately 24.5 L/mol) and the for simplicity envision that \(5 \times 10^{-4} \) moles of CO2 are exhaled every minute. According to the equation, 1 mole of Na2O2 reacts with 2 moles of CO2, so the consumption rate of Na2O2 is \(2.5 \times 10^{-4} \) moles per minute. Multiplying this by the molar mass of Na2O2 (\(77.9778 \,g/mol\)) gives the consumption rate in grams per minute. This should be multiplied by 60 to get the consumption rate in grams per hour.

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is essential for understanding how substances react. In a balanced equation, the number of atoms for each element is the same on both sides of the equation. This ensures that mass is conserved during the reaction because atoms are neither created nor destroyed. For our specific reaction: solid sodium peroxide (\(\text{Na}_2 \text{O}_2\)) reacts with \(\text{CO}_2\) gas to form sodium carbonate (\(\text{Na}_2 \text{CO}_3\)) and oxygen gas (\(\text{O}_2\)). The balanced equation is:\[\text{Na}_2\text{O}_2(\text{s}) + 2\text{CO}_2(\text{g}) \rightarrow 2\text{Na}_2 \text{CO}_3(\text{s}) + \text{O}_2(\text{g})\]In this equation, each side has the same number of sodium, carbon, and oxygen atoms. Balancing equations helps to predict the quantities of reactants and products involved, which is crucial for any calculations linked to stoichiometry.

Gas Volume Calculations

Calculating the volume of gases is a critical step in understanding reactions that involve gaseous substances. In our problem, we need to determine the volume of \(\text{O}_2\) produced per minute. Knowing that the volume of gases exchanged in the lungs is \(4.0\, \text{L/min}\) and that \(3.8\%\) of that is \(\text{CO}_2\), we calculate the volume of \(\text{CO}_2\) as follows:

• \(4.0 \, \text{L/min} \times 0.038 = 0.152 \, \text{L/min} = 152 \, \text{mL/min}\)

From the balanced equation, 1 mole of \(\text{O}_2\) is produced from 2 moles of \(\text{CO}_2\). Therefore, the volume of \(\text{O}_2\) produced per minute is half the volume of \(\text{CO}_2\), giving us \(76 \, \text{mL/min}\). This direct relationship is often used in gas volume calculations when dealing with balanced equations.

Stoichiometry

Stoichiometry involves using relationships from balanced chemical equations to calculate the amounts of reactants and products. It is the backbone of quantitative chemistry. For the given reaction, stoichiometry helps us understand how \(1\) mole of \(\text{Na}_2\text{O}_2\) reacts with \(2\) moles of \(\text{CO}_2\). By applying stoichiometry, we can calculate how much \(\text{Na}_2\text{O}_2\) is consumed based on the \(\text{CO}_2\) volume:

• Determine moles of \(\text{CO}_2\) based on known volume and molar volume
• Molar volume at \(25^{\circ} \text{C}\) and \(735 \, \text{mmHg}\) is approximately \(24.5\, \text{L/mol}\)
• Calculate \(\text{Na}_2\text{O}_2\) consumption: \(2.5 \times 10^{-4}\, \text{moles/min}\)

This molar approach allows us to transition from moles to grams, accommodating conversions for practical measurements.

Molar Volume

Molar volume is the volume one mole of a substance occupies at a given temperature and pressure. For gases at standard conditions (often approximated as \(25^{\circ} \text{C}\) and \(1\, \text{atm}\)), the molar volume is around \(24.5\, \text{L/mol}\). This value is crucial when converting between volumes of gases and quantities in moles.In our calculations, knowing the molar volume allows us to estimate the moles of gas exhaled and needed for reactions:

• Converting from gas volume in liters to moles using molar volume
• Understanding the relationships and stoichiometry in reactions involving gases

By leveraging these conversions, we can readily determine how much of a reactant is consumed, such as \(\text{Na}_2\text{O}_2\), which is pivotal for practical applications like in submarines or space vehicles, where precise measurement is critical.