One method of removing CO2(g) from a spacecraft is to allow the CO2 to react with LiOH.
2LiOH(s)+CO2(g)---->Li2CO3(s)+H2O(l)
How many liters of CO2(g) at 25.5 degrees celsius and 749 Torr can be removed per 9kg of LiOH consumed?
Convert 9 kg LiOH to moles. moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles LiOH to moles CO2.
Now convert moles CO2 to volume using
PV = nRT
To find out how many liters of CO2 can be removed per 9kg of LiOH consumed, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.08206 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's convert the given values to the appropriate units:
Pressure (P) = 749 Torr = 749/760 atm (since 1 atm = 760 Torr)
Temperature (T) = 25.5 °C = 25.5 + 273.15 K
Now, we need to consider the balanced chemical equation given for the reaction between LiOH and CO2:
2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)
From the balanced equation, we can see that 1 mole of CO2 reacts with 2 moles of LiOH.
Next, we need to determine the number of moles of LiOH consumed. Since the molar mass of LiOH is 23.95 g/mol, we can convert the given mass of LiOH (9 kg) to moles:
9 kg LiOH = 9000 g LiOH
Number of moles of LiOH = (9000 g LiOH) / (23.95 g/mol LiOH)
Now, we have the number of moles of LiOH, and since the ratio between LiOH and CO2 is 2:1, we can determine the moles of CO2 consumed:
Moles of CO2 = (1/2) * (Number of moles of LiOH)
Finally, with the moles of CO2, we can calculate the volume using the ideal gas law equation:
V = (nRT) / P
Substituting the values into the equation, we have:
V = [(1/2) * (Number of moles of LiOH) * (0.08206 L·atm/(mol·K)) * (25.5 + 273.15 K)] / (749/760 atm)
Now we can substitute the calculated values from the previous steps and solve for the volume of CO2:
V = [(1/2) * (9000 g LiOH) / (23.95 g/mol LiOH) * (0.08206 L·atm/(mol·K)) * (25.5 + 273.15 K)] / (749/760 atm)
Calculating this expression will give us the volume of CO2 in liters that can be removed per 9 kg of LiOH consumed.
To find out the number of liters of CO2(g) that can be removed per 9 kg of LiOH consumed, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure of CO2(g) in atm
V = volume of CO2(g) in liters
n = number of moles of CO2(g)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
First, we need to convert the pressure from Torr to atm. We can use the following conversion factor:
1 atm = 760 Torr
Therefore, the pressure of CO2(g) in atm is:
749 Torr / 760 Torr/atm = 0.984 atm
Next, we need to convert the temperature from degrees Celsius to Kelvin by adding 273.15:
25.5 degrees Celsius + 273.15 = 298.65 K
Now we can rearrange the ideal gas law equation to solve for the volume (V):
V = (nRT) / P
We already have the values for P, R, and T, but we need to find the number of moles of CO2(g) before we can calculate the volume. The balanced equation tells us that:
2 moles of LiOH react with 1 mole of CO2
Therefore, the number of moles of CO2 per 9 kg of LiOH consumed is:
(9 kg LiOH) / (23.94 g/mol LiOH) * (1 mol CO2) / (2 mol LiOH) = (9 * 1000 g) / (23.94 * 2) = 187.5 mol
Now we can substitute the values into the equation:
V = (187.5 mol * 0.0821 L·atm/mol·K * 298.65 K) / 0.984 atm
Calculating this gives us:
V = 4702.11 L
Therefore, approximately 4702.11 liters of CO2(g) at 25.5 degrees Celsius and 749 Torr can be removed per 9 kg of LiOH consumed.