The space shuttle environment control system handles excess CO2 (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 7 astronauts on board and each exhales 20. L of air per minute, how long could clean air be generated if there were 25,000g of LiOH available. Assume the density of air to be 0.0010 g/ml
first you balance the equation:
CO2+2Li(OH)-->Li2(CO3)+H2O
so it's 20L/min of air per passenger so that's 20*7 gives you 140L/min and then you find 4% of that which is 5.6L/min CO2. then you make a stoich table:
5.6L CO2/min * 1000mL/1 L * 0.0010g CO2/1 ml * 1 mol CO2/44.01g
* 2 mol LiOH/1 mol CO2 * 23.95 g/1 mol LiOH * 1/25000g
so then you multiply all the top and divide by all of the bottom and get: 2.4e-4 min^-1 (minutes are on the bottom) so just inverse that and you get 4108 minutes and that should be the answer.
This is about 2.85 days or about 2 days 20 hours 18 minutes and 21 seconds.
Hydroxonitrat aci
To find out how long clean air can be generated, we need to calculate the total amount of CO2 that can be processed by the given amount of LiOH.
1. Calculate the mass of CO2 produced by one astronaut per minute:
Exhaled air volume per minute = 20 L/minute
Density of air = 0.0010 g/ml
Mass of exhaled air per minute = (20 L/minute) * (0.0010 g/ml) = 0.02 kg/minute
Mass of CO2 in exhaled air per minute = 0.02 kg/minute * 0.040 = 0.0008 kg/minute
2. Calculate the total mass of CO2 produced by all the astronauts per minute:
Total mass of CO2 per minute = 7 astronauts * 0.0008 kg/minute = 0.0056 kg/minute
3. Calculate the total mass of CO2 that can be processed by the given amount of LiOH:
Mass of LiOH available = 25000 g = 25 kg
The balanced chemical equation for the reaction is:
2 LiOH + CO2 → Li2CO3 + H2O
From the equation, we know that 2 moles of LiOH react with 1 mole of CO2.
The molar mass of LiOH = 23.95 g/mol + 16.00 g/mol + 1.01 g/mol = 40.96 g/mol
The molar mass of CO2 = 12.01 g/mol + (2 * 16.00 g/mol) = 44.01 g/mol
So, 25 kg of LiOH is equivalent to (25000 g / 40.96 g/mol) = 610.03 moles of LiOH.
Therefore, the total mass of CO2 that can be processed by the given amount of LiOH is (610.03 moles * 1 mole of CO2 / 2 moles of LiOH * 44.01 g/mol) = 13,437.44 g = 13.44 kg
4. Divide the total mass of CO2 by the rate of CO2 production to find the time duration:
Time duration = 13.44 kg / 0.0056 kg/minute ≈ 2400 minutes
Therefore, clean air can be generated for approximately 2400 minutes based on the given amount of LiOH available.
To find out how long clean air could be generated, we need to calculate the amount of CO2 produced by the astronauts and then determine how much LiOH is required to react with that amount of CO2.
First, let's calculate the amount of CO2 produced by the astronauts per minute:
CO2 produced per minute = 7 astronauts * 20. L/airman * 0.0010 g/ml * 4.0% CO2 = 0.56 g CO2/minute
Now, we need to determine the stoichiometry of the reaction between CO2 and LiOH to find out how much LiOH is required to react with 0.56 g of CO2:
The balanced chemical equation for the reaction is:
2 LiOH + CO2 → Li2CO3 + H2O
From the equation, we can see that 2 moles of LiOH react with 1 mole of CO2. Now, let's calculate the molar mass of CO2:
Molar mass of CO2 = 1 * atomic mass of C + 2 * atomic mass of O
= 1 * 12.01 g/mol + 2 * 16.00 g/mol
= 44.01 g/mol
Using the molar mass of CO2, we can calculate the number of moles of CO2 produced per minute:
Number of moles of CO2 produced per minute = 0.56 g CO2/minute / 44.01 g/mol
≈ 0.0127 mol CO2/minute
Since the stoichiometry of the reaction is 2 moles of LiOH to 1 mole of CO2, we need twice the number of moles of LiOH to react with CO2:
Number of moles of LiOH required per minute = 2 * 0.0127 mol CO2/minute
≈ 0.0254 mol LiOH/minute
Finally, we can calculate the total time clean air could be generated with the given amount of LiOH:
Total time = Amount of LiOH available / Number of moles of LiOH required per minute
= 25000 g / 0.0254 mol/minute
≈ 984252 minutes
So, clean air could be generated for approximately 984,252 minutes with the given amount of LiOH.