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Chapter 3: Problem 120

The space shuttle environmental control system handled 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, Li \(_{2} \mathrm{CO}_{3},\) and water. If there were 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 \(\mathrm{g} / \mathrm{mL}\) .

Short Answer

Expert verified
Clean air could be generated for approximately 4103.09 minutes, or about 68 hours and 23 minutes, using 25,000 g of LiOH pellets for the given shuttle mission.

Step by step solution

01

Calculate the total mass of exhaled air by the astronauts per minute

Let's find the total mass of exhaled air per minute by all seven astronauts. As the density of air is 0.0010 g/mL and each astronaut exhales 20 L of air per minute, we can calculate the mass as follows: Total volume of exhaled air by 7 astronauts = 7 astronauts * 20 L/minute * 1000 mL/L = 140,000 mL/minute Mass of exhaled air = Volume * Density = 140,000 mL/minute * 0.0010 g/mL = 140 g/minute
02

Calculate the total mass of exhaled CO2 by the astronauts per minute

Given that the CO2 comprises 4% of the mass of exhaled air, we can calculate the mass of CO2 exhaled per minute as: Mass of CO2 = 4% of Mass of exhaled air per minute = (4/100) * 140 g/minute = 5.6 g/minute
03

Write the balanced chemical equation for the reaction

We are given that the CO2 reacts with LiOH to form Li2CO3 and water. Writing and balancing the chemical equation for the reaction: 2 LiOH + CO2 → Li2CO3 + H2O
04

Determine the mass of CO2 that reacts with the given mass of LiOH

First, we need to find the moles of LiOH, given that there are 25,000 g of LiOH pellets available. The molar mass of LiOH = 6.94 (Li) + 15.999 (O) + 1.007 (H) = 23.95 g/mol. Moles of LiOH = Mass / Molar mass = 25,000 g / 23.95 g/mol = 1043.84 mol Now, from the balanced equation, 2 moles of LiOH react with 1 mole of CO2. Therefore, the moles of CO2 that react with 1043.84 moles of LiOH = 1043.84/2 = 521.92 moles. The molar mass of CO2 = 12.01 (C) + 15.999 x 2 (O) = 44.009 g/mol. Mass of CO2 that can react with 25,000 g of LiOH: = 521.92 mol * 44.009 g/mol = 22,977.32 g
05

Calculate the duration for which the LiOH pellets can generate clean air

Now that we know the mass of CO2 that can react with the given mass of LiOH and the mass of CO2 exhaled by astronauts per minute, we can calculate the duration for which clean air can be generated as: Duration = (Mass of CO2 that can react with LiOH) / (Mass of CO2 exhaled per minute) = 22,977.32 g / 5.6 g/minute = 4103.09 minutes Therefore, clean air could be generated for approximately 4103.09 minutes, or about 68 hours and 23 minutes, using 25,000 g of LiOH pellets for the given shuttle mission.

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