The space shuttle environmental control system handles excess CO2 (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, Li2CO3 , and water. If there are four astronauts on board the shuttle, and each exhales 20. L of air per minute, how long could clean air be generated if there were 28,000 g of LiOH pellets available for each shuttle mission? Assume the density of air is 0.0010 g/mL.

Question

The space shuttle environmental control system handles excess CO2 (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, Li2CO3 , and water. If there are four astronauts on board the shuttle, and each exhales 20. L of air per minute, how long could clean air be generated if there were 28,000 g of LiOH pellets available for each shuttle mission? Assume the density of air is 0.0010 g/mL.

Answer 1

To solve this problem, we need to calculate the amount of CO2 produced by the astronauts and the amount of LiOH that reacts with the CO2.

1. Calculate the amount of CO2 produced by the astronauts:
- Each astronaut exhales 20 L of air per minute.
- Since CO2 makes up 4.0% by mass of exhaled air, the mass of CO2 exhaled by each astronaut per minute is: 20 L * 0.0010 g/mL * 4.0% = 0.08 g

2. Calculate the total amount of CO2 produced by all four astronauts:
- Since there are four astronauts aboard the shuttle, the total amount of CO2 produced per minute is: 0.08 g/athlete * 4 athletes = 0.32 g/min

3. Calculate the amount of LiOH required to react with the CO2:
- The balanced chemical equation for the reaction is:
2LiOH + CO2 → Li2CO3 + H2O
From the equation, we can see that 2 moles of LiOH react with 1 mole of CO2.
- The molar mass of LiOH is: 6.941 g/mol + 16.00 g/mol + 1.008 g/mol = 23.95 g/mol
- The molar mass of CO2 is: 12.01 g/mol + 2 * 16.00 g/mol = 44.01 g/mol
- Using the molar mass, we can calculate the amount of LiOH needed in grams: (0.32 g CO2/min) * (2 mol LiOH / 1 mol CO2) * (23.95 g/mol LiOH) / (44.01 g/mol CO2) = 0.348 g/min

4. Determine the duration of clean air production:
- If there are 28,000 g of LiOH pellets available, we can calculate the time it takes to consume all the LiOH: 28,000 g / (0.348 g/min) = 80,460 min
- Converting the time to hours: 80,460 min / 60 min/hour = 1,341 hours

Therefore, if there are 28,000 g of LiOH pellets available for each shuttle mission, clean air could be generated for approximately 1,341 hours.

Answer 2

This is an example of a "work horse" problem designed to make you think through many steps. Do what you can, show your work, and someone will help you finish if you explain what you don't understand about the next step.