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!