LiOH and NaOH can each react with CO2
to form the metal carbonate and H2O.
These reactions can be used to remove CO2
from the air in a spacecraft.
a. Write a balanced equation for each
reaction.
b.In order to remove 214.3 g CO2 from the air, what mass of NaOH would be consumed? What mass of LiOH would be required to remove the same amount of CO2?
c. Based on your calculations, explain why LiOH is used during space missions rather than NaOH even though LiOH can cost up to 20x more than the NaOH per gram. Hint: SpaceX is hoping to get the cost of launching material to space down to “only” $1000/pound.
a. 2LiOH + CO2 ==> Li2CO3 + H2O
2NaOH + CO2 --> Na2CO3 + H2O
b. mols CO2 = grams/molar mass = 214.3/44 = 4.87
Convert to mols NaOH. 4.87 mols CO2 x (2 mols NaOH/1 mol CO2) = 9.74 mols NaOH needed. mass NaOH = mols NaOH x molar mass NaOH 9.74 x 40 = about 390 g but that's an estimate. You do it more accurately.
Same process for LiOH. Mols is same so 9.74 x 23.9 = estimate 233 g.
c. mols needed is same for LiOH and NaOH but the molar mass is so much lower for LiOH than NaOH that it requires much less LiOH than NaOH to remove the same amount of CO2. That weight difference means saving a lot of fuel to transport and launch the rocket.
a. The balanced equations for the reactions are:
1. LiOH + CO2 → Li2CO3 + H2O
2. 2 NaOH + CO2 → Na2CO3 + H2O
b. To find the mass of NaOH required to remove 214.3 g of CO2, we need to use the stoichiometry of the reaction (in equation 2) to determine the mole ratio.
From the equation, we can see that 2 moles of NaOH react with 1 mole of CO2. Using the molar mass of CO2 (44.01 g/mol), we can convert the mass of CO2 to moles:
Moles of CO2 = (Mass of CO2 / Molar mass of CO2)
Moles of CO2 = (214.3 g / 44.01 g/mol) ≈ 4.868 mol CO2
Since the mole ratio is 2:1 (NaOH:CO2), we can calculate the moles of NaOH required:
Moles of NaOH = 2 × Moles of CO2
Moles of NaOH = 2 × 4.868 mol ≈ 9.736 mol NaOH
Then, using the molar mass of NaOH (39.997 g/mol), we can convert the moles of NaOH to grams:
Mass of NaOH = Moles of NaOH × Molar mass of NaOH
Mass of NaOH = 9.736 mol × 39.997 g/mol ≈ 389.38 g NaOH
Therefore, approximately 389.38 g of NaOH would be consumed to remove 214.3 g of CO2.
For LiOH, the balanced equation tells us that the mole ratio is 1:1 (LiOH:CO2). So, the same amount of moles of LiOH would be required:
Moles of LiOH = Moles of CO2
Moles of LiOH = 4.868 mol LiOH (same as CO2)
Using the molar mass of LiOH (23.95 g/mol), we can calculate the mass of LiOH:
Mass of LiOH = Moles of LiOH × Molar mass of LiOH
Mass of LiOH = 4.868 mol × 23.95 g/mol ≈ 116.6 g LiOH
Therefore, approximately 116.6 g of LiOH would be required to remove the same amount of CO2.
c. The cost consideration for space missions is crucial. Although LiOH is more expensive per gram compared to NaOH, its lighter atomic weight makes a significant difference when it comes to the cost of launching materials to space. Because every additional gram adds to the total payload weight, using LiOH, which weighs less than NaOH per mole, could be more cost-effective in terms of launch expenses. Hence, despite the higher cost per gram, LiOH may still be preferred during space missions.
a. The balanced equations for the reactions are:
1. LiOH + CO2 → Li2CO3 + H2O
2. NaOH + CO2 → Na2CO3 + H2O
b. To determine the mass of NaOH consumed, we need to know the molar mass of CO2 and the molar mass of NaOH.
Molar mass of CO2 = 12.01 g/mol + (16.00 g/mol x 2) = 44.01 g/mol
Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Using the molar mass, we can calculate the number of moles of CO2:
Number of moles of CO2 = 214.3 g / 44.01 g/mol = 4.87 mol
Since the balanced equation shows a 1:1 stoichiometric ratio between CO2 and NaOH, the same number of moles of NaOH will be consumed:
Number of moles of NaOH = 4.87 mol
Finally, to determine the mass of NaOH consumed:
Mass of NaOH = Number of moles of NaOH × Molar mass of NaOH
Mass of NaOH = 4.87 mol × 40.00 g/mol = 194.8 g
For LiOH, we use the same process. The molar mass of LiOH is:
Molar mass of LiOH = 6.94 g/mol + 16.00 g/mol + 1.01 g/mol = 23.95 g/mol
Number of moles of LiOH = 4.87 mol
Mass of LiOH consumed = Number of moles of LiOH × Molar mass of LiOH
Mass of LiOH consumed = 4.87 mol × 23.95 g/mol = 116.5 g
c. LiOH is used during space missions instead of NaOH, even though it can be more expensive per gram, because the cost of launching material into space is extremely high. LiOH is chosen due to its higher molar mass compared to NaOH (23.95 g/mol vs. 40.00 g/mol). This means that a smaller mass of LiOH can achieve the same stoichiometric conversion as a larger mass of NaOH. Using LiOH allows for a reduction in the amount of reagent that needs to be transported to space, resulting in lower transportation costs.
a. To write the balanced equations, we first need to identify the chemical formulas of the reactants and products in the reaction:
LiOH (Lithium hydroxide) reacts with CO2 (Carbon dioxide) to form Li2CO3 (Lithium carbonate) and H2O (Water).
The balanced equation for the reaction is:
2 LiOH + CO2 → Li2CO3 + H2O
NaOH (Sodium hydroxide) also reacts with CO2 to form Na2CO3 (Sodium carbonate) and H2O.
The balanced equation for the reaction is:
2 NaOH + CO2 → Na2CO3 + H2O
b. To calculate the mass of NaOH that would be consumed to remove 214.3 g of CO2, we need to use the stoichiometry of the balanced equation.
From the balanced equation:
2 NaOH + CO2 → Na2CO3 + H2O
We can see that 2 moles of NaOH react with 1 mole of CO2. The molar mass of CO2 is 44.01 g/mol.
Now we can set up a proportion to find the mass of NaOH:
(2 moles NaOH / 1 mole CO2) = (x g NaOH / 214.3 g CO2)
Solving for x (mass of NaOH):
x = (2 moles NaOH / 1 mole CO2) * (214.3 g CO2)
Therefore, the mass of NaOH required to remove 214.3 g of CO2 would be approximately (2 * 214.3) g = 428.6 g.
To find the mass of LiOH required to remove the same amount of CO2, we use the same method and stoichiometry as above.
c. Based on the calculations, it might seem counterintuitive to use LiOH instead of NaOH despite its higher cost. However, there could be other factors to consider in space missions. LiOH is chosen over NaOH primarily because of its lower molecular weight. Launching material into space is expensive, and reducing the weight of the spacecraft is crucial for cost-effective missions. LiOH has a lower molecular weight compared to NaOH, meaning less mass is required to achieve the same CO2 removal. Hence, even though LiOH may be more expensive per gram, the reduced amount required for the same CO2 removal justifies its usage in space missions where weight and cost are major considerations.