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Chapter 3: Problem 106

The space shuttle environmental control system handles excess \(\mathrm{CO}_{2}\) (which the astronauts breathe out; it is \(4.0 \%\) 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 7 astronauts on board the shuttle, and each exhales \(20 . \mathrm{L}\) of air per minute, how long could clean air be generated if there were \(25,000 \mathrm{~g}\) of LiOH pellets available for each shuttle mission? Assume the density of air is \(0.0010 \mathrm{~g} / \mathrm{mL}\).

Short Answer

Expert verified
The 25,000 g of LiOH pellets can generate clean air for approximately 4110 minutes if there are 7 astronauts on board.

Step by step solution

01

Calculating the mass of \(\mathrm{CO}_{2}\) exhaled by one astronaut per minute

We know each astronaut exhales 20 L of air per minute and \(\mathrm{CO}_{2}\) is 4.0% by mass of the exhaled air. We are also given the density of air as 0.0010 g/mL. First, we need to find the mass of the exhaled air and then calculate the mass of \(\mathrm{CO}_{2}\) in that air. Mass of exhaled air per minute by one astronaut = Volume 脳 Density = 20 L 脳 1000 mL/L 脳 0.0010 g/mL = 20 g Now, let's calculate the mass of \(\mathrm{CO}_{2}\) exhaled by one astronaut per minute: Mass of \(\mathrm{CO}_{2}\) = (Mass of exhaled air 脳 Percentage of \(\mathrm{CO}_{2}\)) / 100 = (20 g 脳 4.0) / 100 = 0.8 g
02

Calculating the mass of \(\mathrm{CO}_{2}\) exhaled by all 7 astronauts per minute

Since there are 7 astronauts on board, we need to calculate the total mass of \(\mathrm{CO}_{2}\) exhaled by all of them per minute: Total mass of \(\mathrm{CO}_{2}\) per minute = Mass of \(\mathrm{CO}_{2}\) per astronaut 脳 Number of astronauts = 0.8 g 脳 7 = 5.6 g
03

Calculate the moles of \(\mathrm{CO}_{2}\) and LiOH

Next, we need to calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled per minute and the moles of LiOH provided: Moles of \(\mathrm{CO}_{2}\) per minute \(= \frac{\text{mass of}\ \mathrm{CO}_{2}\ \text{per minute}}{\text{molar mass of}\ \mathrm{CO}_{2}}\) Moles of \(\mathrm{CO}_{2}\) per minute \(= \frac{5.6\ g}{(12.01\ g/mol\ \text{for C}) + (2 \times 16.00\ g/mol\ \text{for O})} = \frac{5.6\ g}{44.01\ g/mol} \approx 0.127\) mol Moles of LiOH \(= \frac{\text{mass of LiOH}}{\text{molar mass of LiOH}}\) Moles of LiOH \(= \frac{25000 g}{(6.94\ g/mol\ \text{for Li}) + (15.99\ g/mol\ \text{for O}) + (1.008\ g/mol\ \text{for H})} = \frac{25000\ g}{23.948\ g/mol} \approx 1044\) mol
04

Calculate the theoretical time to generate clean air

Using the balanced chemical equation, \(2\text{LiOH} + \mathrm{CO}_{2} \rightarrow \mathrm{Li}_{2}\mathrm{CO}_{3} + \text{H}_{2}\text{O}\), we can deduce that 2 moles of LiOH reacts with 1 mole of \(\mathrm{CO}_{2}\). Therefore, let's find how many moles of \(\mathrm{CO}_{2}\) can react with the available LiOH: Moles of \(\mathrm{CO}_{2}\) reacting with the available LiOH = \(\frac{1}{2} \times\) Moles of LiOH = \(\frac{1}{2} \times 1044 \text{ moles} = 522\) mol Now, let's calculate how many minutes this amount of LiOH can generate clean air, considering the moles of \(\mathrm{CO}_{2}\) exhaled per minute: Minutes to generate clean air \(= \frac{\text{total moles of}\ \mathrm{CO}_{2}\ \text{reacting with LiOH}}{\text{moles of}\ \mathrm{CO}_{2}\ \text{exhaled per minute}}\) Minutes to generate clean air \(= \frac{522 \text{ mol}}{0.127 \text{ mol/min}} \approx 4110\) minutes Hence, the 25,000 g of LiOH pellets can generate clean air for approximately 4110 minutes if there are 7 astronauts on board.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carbon Dioxide Removal
In a space shuttle, managing the levels of carbon dioxide (CO鈧�) is crucial to ensure a safe environment for the astronauts. As astronauts breathe, they release CO鈧� into the cabin air. If left unchecked, elevated CO鈧� levels can be harmful.

To remove CO鈧�, the environmental control system uses chemical reactions to transform the gas into a less harmful substance. This process involves capturing CO鈧� before it builds up to dangerous levels.

Unlike ventilation in homes here on Earth, space shuttles rely on chemical processes due to the closed environment and present challenges. Thus, CO鈧� removal is an integral part of life support systems in space missions.
Lithium Hydroxide Reaction
Lithium hydroxide (LiOH) is a crucial component used in the chemical process to remove CO鈧� from a space shuttle. The reaction involves lithium hydroxide reacting with carbon dioxide to form lithium carbonate and water. This is an efficient process for CO鈧� removal.

The balanced chemical equation for this reaction is: 2 LiOH + CO鈧� 鈫� Li鈧侰O鈧� + H鈧侽. This equation indicates that two moles of LiOH react with one mole of carbon dioxide. The compound lithium carbonate and water produced are non-toxic, which makes this chemical reaction particularly suitable for space missions.

The substitution of harmful CO鈧� with harmless by-products helps maintain breathable air on board. Given the constrained space and resources, using lithium hydroxide is a practical solution for maintaining air quality.
Astronauts' Exhaled Air
Inside a space shuttle, astronauts exhale a significant volume of air filled with 4% carbon dioxide by mass. Given that each astronaut exhales around 20 liters of air per minute, managing this exhaled air is essential.

Understanding the properties of exhaled air is necessary for efficient air recycling and planning for long-term space missions. With a known density of air (0.0010 g/mL), calculations can determine the mass of CO鈧� exhaled, aiding in proper system design.

Managing exhaled air is not only about keeping the air clean but also conserving resources on prolonged missions. Efficiently converting exhaled air into breathable form enables astronauts to stay longer in space without external replenishment.
Chemical Stoichiometry
Chemical stoichiometry plays a pivotal role in determining the exact amounts of substances involved in chemical reactions. In the context of removing CO鈧� from spacecraft, stoichiometry is vital to calculate how much lithium hydroxide is required to react with the CO鈧� exhaled by astronauts.

The stoichiometric calculations involve determining the molar amounts from the mass of the substances involved. In the given exercise, the molar mass of CO鈧� and LiOH are crucial to deducing the efficiency of the reaction.

By calculating moles and using balanced equations, one can determine precisely how long the available LiOH can sustain clean air levels. This ensures that space mission planners can make accurate predictions for the duration of missions, considering all vital resources and reaction rates.

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Most popular questions from this chapter

Arrange the following substances in order of increasing mass percent of carbon. a. caffeine, \(\mathrm{C}_{\mathrm{s}} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\) b. sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) c. ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of \(0.157 \mathrm{~g}\) of the compound produced \(0.213 \mathrm{~g}\) \(\mathrm{CO}\), and \(0.0310 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that \(0.103 \mathrm{~g}\) of the compound produces \(0.0230 \mathrm{~g} \mathrm{NH}_{3} .\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

A common demonstration in chemistry courses involves adding a tiny speck of manganese(IV) oxide to a concentrated hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) solution. Hydrogen peroxide decomposes quite spectacularly under these conditions to produce oxygen gas and steam (water vapor). Manganese(IV) oxide is a catalyst for the decomposition of hydrogen peroxide and is not consumed in the reaction. Write the balanced equation for the decomposition reaction of hydrogen peroxide.

A \(2.077-g\) sample of an element, which has an atomic mass between 40 and 55 , reacts with oxygen to form \(3.708 \mathrm{~g}\) of an oxide. Determine the formula of the oxide (and identify the element).

Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \% \mathrm{H} \text { (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \text { II } & 12.58 & 87.42 \\ \text { III } & 2.34 & 97.66 \end{array} $$ When \(1.00 \mathrm{~L}\) of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$ \begin{array}{lcc} & \mathrm{H}_{2} \text { (L) } & \mathrm{N}_{2} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \text { II } & 2.00 & 1.00 \\ \text { III } & 0.50 & 1.50 \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.
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