Question

The space shuttle environmental control system handles excess ${\mathrm{CO}}_2$ (which the astronauts breathe out; it is $4.0\mathrm{\%}$ by mass of exhaled air) by reacting it with lithium hydroxide, LiOH, pellets to form lithium carbonate, ${\mathrm{Li}}_2{\mathrm{CO}}_3$, and water. If there are seven astronauts on board the shuttle, and each exhales 20. L of air per minute, how long could clean air be generated if there were 25,000 g of LiOH pellets available for each shuttle mission? Assume the density of air is 0.0010 g/mL.

Short answer

The available LiOH can generate clean air for approximately 68.5 hours for seven astronauts.

Step-by-step solution

Calculate the mass of CO₂ exhaled per minute

Firstly, we need to determine the mass of CO₂ exhaled by seven astronauts each minute. To do this, we will use the given information that CO₂ comprises 4.0% by mass of the exhaled air and the density of air is 0.0010 g/mL. The volume of air exhaled per minute by one astronaut is 20 L, which is equivalent to 20,000 mL: Mass of air exhaled per minute by one astronaut = Volume x Density = 20,000 mL x 0.0010 g/mL = 20 g As there are seven astronauts, the total mass of air exhaled per minute is 7 x 20 g = 140 g. Since CO₂ is 4.0% by mass of the exhaled air: Mass of CO₂ exhaled per minute by seven astronauts = 0.040 x 140 g = 5.6 g

Write the balanced chemical equation

The balanced chemical equation representing the reaction between CO₂ and LiOH to form Li₂CO₃ and H₂O is: 2 LiOH + CO₂ → Li₂CO₃ + H₂O

Use stoichiometry to find the mass of LiOH required

From the balanced equation, we can see that 2 moles of LiOH react with 1 mole of CO₂. To find out how much LiOH is required to react with the CO₂ exhaled per minute, we need to convert the mass of CO₂ to moles and then use the stoichiometry to find the mass of LiOH: Molar mass of CO₂ = 12.01 g/mol (C) + 2 x 16.00 g/mol (O) = 44.01 g/mol Moles of CO₂ exhaled per minute = Mass / Molar Mass = 5.6 g / 44.01 g/mol ≈ 0.127 mol Now, use the balanced equation stoichiometry (2 moles of LiOH react with 1 mole of CO₂), we can find the moles of LiOH required: Moles of LiOH required = 2 x 0.127 mol ≈ 0.254 mol To find the mass of LiOH, we use the molar mass of LiOH (6.94 g/mol (Li) + 15.99 g/mol (O) + 1.01 g/mol (H) = 23.94 g/mol): Mass of LiOH required per minute = Moles x Molar Mass = 0.254 mol x 23.94 g/mol ≈ 6.08 g

Calculate how long the LiOH will last

We have been given that there are 25,000 g of LiOH pellets available for each shuttle mission. To find out how long it would last, divide the total available mass of LiOH by the mass required per minute: Duration (minutes) = Total mass of LiOH / Mass of LiOH required per minute = 25,000 g / 6.08 g/min ≈ 4,111 min To convert minutes to hours, divide the result by 60: Duration (hours) = 4,111 min / 60 min/h ≈ 68.5 h Thus, the available LiOH can generate clean air for approximately 68.5 hours for seven astronauts.

Key concepts

Balanced Chemical Equation

A balanced chemical equation is essential for understanding chemical reactions. It ensures that the same number of each type of atom appears on both sides of the equation. This adheres to the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
The given exercise involves the reaction between carbon dioxide (CO₂) and lithium hydroxide (LiOH) to form lithium carbonate (Li₂CO₃) and water (H₂O). The balanced equation for this reaction is:

• 2 LiOH + CO₂ → Li₂CO₃ + H₂O

Here, the balance is achieved as follows:

• 2 lithium (Li) atoms on both sides
• 2 oxygen (O) atoms from LiOH, plus one from CO₂, equalling three oxygen atoms on both sides
• 1 carbon (C) atom from CO₂, creating balance with one carbon atom in Li₂CO₃
• 2 hydrogen (H) atoms in total on both sides, from H₂O

Balanced equations are crucial for correctly using stoichiometry in chemistry.

Molar Mass

Molar mass is the mass of one mole of a substance measured in grams per mole (g/mol). It is crucial when converting between the mass of a substance and the number of moles, which is a foundational concept in stoichiometry. Here’s how we calculate it for the substances in the exercise:

• Carbon dioxide (CO₂) has a molar mass calculated as the sum of its atomic masses: 12.01 g/mol for carbon (C) plus 2 x 16.00 g/mol for oxygen (O), totaling 44.01 g/mol.
• Lithium hydroxide (LiOH) has a molar mass calculated from: 6.94 g/mol for lithium (Li), 15.99 g/mol for oxygen (O), and 1.01 g/mol for hydrogen (H), totaling 23.94 g/mol.

Understanding molar mass allows you to determine how many molecules or ions you have in a given mass of a substance. This is crucial for stoichiometric calculations, like determining how much reagent is needed for a reaction.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. Each reaction has a balanced equation, which reveals the proportions of each reactant and product involved. In the provided exercise, we see an important chemical reaction between carbon dioxide and lithium hydroxide.

• Reactants: CO₂ and LiOH
• Products: Li₂CO₃ and H₂O

Chemical reactions can only proceed as long as there is a sufficient amount of reactants. Understanding the stoichiometry—the quantitative relationships implied by the balanced chemical equation—is essential.

• For every mole of CO₂, two moles of LiOH are required.

This 1:2 stoichiometric ratio in the reaction indicates how one reactant may run out before others, dictating the extent and duration of the reaction.

Gas Laws

Gas laws describe the behavior of gases under various conditions of pressure, volume, temperature, and amount. Though not explicitly used in this exercise, understanding the concepts is vital.
For example, the Ideal Gas Law combines several simple gas laws into one equation:

• PV = nRT

Where

• P is pressure,
• V is volume,
• n is the number of moles,
• R is the gas constant,
• T is the temperature.

In the exercise, understanding that the gas volume (like the volume of CO₂ exhaled) changes with conditions and the stoichiometry of the reaction can help further conceptualize the process.

• Recognizing how gas density dimensions are used (e.g., 0.0010 g/mL in the exercise) can assist in calculating masses.

These principles help predict how changing one property (like volume) affects others.