One method of removing CO2(g) from a spacecraft is to allow the CO2 to react with LiOH. What volume, in liters, of CO2(g) at 26.0°C and 793 Torr can be removed per kilogram of LiOH consumed?
2LiOH(s) + CO2(g) -> Li2CO3(s) + H2O(l)
Use stoichiometry to convert 1 kg LiOH to liters CO2 at STP. Convert liters CO2 at STP to mols, then use PV=nRT to adjust to 26 C and 793 Torr. Post your work if you get stuck.
Oh, spacecrafts and chemical reactions? Sounds like a recipe for a good laugh!
Alright, let's get started. We have the chemical equation:
2LiOH(s) + CO2(g) -> Li2CO3(s) + H2O(l)
And we want to find the volume of CO2(g) that can be removed per kilogram of LiOH consumed.
First, we need to convert 1 kg of LiOH to liters of CO2 at STP. But wait, what's STP? Is that an abbreviation for "Seriously Tickling Penguins"?
Anyways, back to business. 1 mole of any gas occupies 22.4 liters at STP. So, we need to know the molar mass of LiOH to calculate the number of moles.
Let's see... the molar mass of LiOH is approximately 23 + 16 + 1 = 40 g/mol. Since we have 1 kg of LiOH, we have 1000 grams of LiOH. Dividing that by the molar mass, we get 1000g / 40 g/mol = 25 moles of LiOH.
Now, according to the balanced equation, we have a 2:1 ratio between LiOH and CO2. So, 25 moles of LiOH will react with 12.5 moles of CO2.
But hold on, we're not done yet. We need to adjust this to the given temperature and pressure of 26.0°C and 793 Torr. So, let's use the gas law equation PV = nRT.
We know the pressure (793 Torr), the temperature (26.0°C = 299.15 K), and the number of moles (12.5 moles). The gas constant R is approximately 0.0821 L•atm/mol•K. So, we can rearrange the equation to solve for the volume V:
V = (nRT) / P
V = (12.5 mol * 0.0821 L•atm/mol•K * 299.15 K) / 793 torr
Now, if you do the math, you will get the volume of CO2 in liters. And there you have it, the volume of CO2 that can be removed per kilogram of LiOH consumed. Happy calculations!
To determine the volume of CO2 that can be removed per kilogram of LiOH consumed, we need to follow a series of steps:
Step 1: Convert 1 kilogram of LiOH to moles of LiOH.
First, we need to determine the molar mass of LiOH:
Li: 1 atomic mass unit (amu)
O: 16 amu
H: 1 amu
Molar mass of LiOH = (1 * 6.941 amu) + (1 * 15.999 amu) + (1 * 1.008 amu) = 23.919 amu
Next, we can convert 1 kilogram (1000 grams) of LiOH to moles using the molar mass:
1 kg LiOH * (1000 g / 23.919 g/mol) ≈ 41.8 moles of LiOH
Step 2: Use the stoichiometry of the reaction to determine moles of CO2 produced.
From the balanced chemical equation, we can see that the stoichiometric ratio between LiOH and CO2 is 2:1. So, for every 2 moles of LiOH consumed, 1 mole of CO2 is produced.
Therefore, 41.8 moles of LiOH * (1 mole CO2 / 2 moles LiOH) = 20.9 moles of CO2
Step 3: Convert moles of CO2 to volume at standard temperature and pressure (STP).
At STP, 1 mole of any ideal gas occupies 22.4 liters of volume.
20.9 moles of CO2 * (22.4 liters CO2 / 1 mole CO2) = 467.36 liters CO2 at STP
Step 4: Adjust the volume of CO2 to 26.0°C and 793 Torr using the Ideal Gas Law (PV = nRT).
First, let's convert 793 Torr to atm:
793 Torr * (1 atm / 760 Torr) ≈ 1.04 atm
Now, we can calculate the volume of CO2 at the given temperature and pressure:
V = (nRT) / P
Plugging in the values:
V = (20.9 moles CO2 * 0.0821 L * atm/mol * K * 299.15 K) / 1.04 atm ≈ 475.66 liters CO2 at 26.0°C and 793 Torr
Therefore, approximately 475.66 liters of CO2(g) at 26.0°C and 793 Torr can be removed per kilogram of LiOH consumed.