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. Calculate the grams of NaOH and of
LiOH that remove 288 g CO2 from the air.
c. NaOH is less expensive per mole than
LiOH. Based on your calculations,
explain why LiOH is used during shuttle
missions rather than NaOH.
thank you so much!
2NaOH + CO2 ==> Na2CO3 + H2O
Convert 288 g CO2 to moles. moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles CO2 to moles NaOH.
Convert moles NaOH to grams. g = moles x molar mass.
You should be able to do the rest on your own. Post your work if you get stuck.
a. Balanced equations:
1. LiOH + CO2 -> Li2CO3 + H2O
2. NaOH + CO2 -> Na2CO3 + H2O
b. To calculate the grams of NaOH and LiOH required to remove 288 g CO2, we need to determine the mole ratios between CO2, NaOH, and LiOH in the balanced equations.
The molar masses of CO2, NaOH, and LiOH are:
CO2: 44.01 g/mol
NaOH: 39.997 g/mol
LiOH: 23.949 g/mol
Using the molar ratios from the balanced equations:
1. For LiOH: 1 mol CO2 reacts with 1 mol LiOH
2. For NaOH: 1 mol CO2 reacts with 1 mol NaOH
So, to calculate the grams of NaOH and LiOH required, we can use the following equations:
Grams of NaOH = (Moles of NaOH / Moles of CO2) * Molar mass of NaOH
Grams of LiOH = (Moles of LiOH / Moles of CO2) * Molar mass of LiOH
For 288 g CO2:
Moles of CO2 = (288 g CO2) / (44.01 g/mol CO2)
For NaOH:
Moles of NaOH = Moles of CO2
Grams of NaOH = (Moles of NaOH / Moles of CO2) * Molar mass of NaOH
For LiOH:
Moles of LiOH = Moles of CO2
Grams of LiOH = (Moles of LiOH / Moles of CO2) * Molar mass of LiOH
Let's calculate the values.
Moles of CO2 = (288 g CO2) / (44.01 g/mol CO2) = 6.54461 mol CO2
Grams of NaOH = (6.54461 mol NaOH / 6.54461 mol CO2) * 39.997 g/mol NaOH
Grams of NaOH = 39.997 g NaOH
Grams of LiOH = (6.54461 mol LiOH / 6.54461 mol CO2) * 23.949 g/mol LiOH
Grams of LiOH = 23.949 g LiOH
c. Although NaOH is less expensive per mole than LiOH, LiOH is used during shuttle missions rather than NaOH because LiOH has a higher absorption capacity for CO2. It can remove a greater amount of CO2 from the air compared to the same mass of NaOH. This makes it more efficient for removing CO2 in the confined space of a spacecraft. So, despite being more expensive, LiOH is preferred due to its superior performance in CO2 removal.
a. The balanced equation for the reaction between LiOH and CO2 is:
2 LiOH + CO2 -> Li2CO3 + H2O
The balanced equation for the reaction between NaOH and CO2 is:
2 NaOH + CO2 -> Na2CO3 + H2O
b. To calculate the grams of NaOH and LiOH needed to remove 288 g of CO2 from the air, we need to use stoichiometry and the molar masses of the compounds involved.
First, let's calculate how many moles of CO2 are present in 288 g. The molar mass of CO2 is 44.01 g/mol.
Moles of CO2 = Mass of CO2 / Molar mass of CO2
Moles of CO2 = 288 g / 44.01 g/mol
Moles of CO2 = 6.546 mol
Since both reactions have a 1:1 stoichiometric ratio between CO2 and the metal carbonate, we know that 6.546 moles of LiOH and NaOH are needed to react with the 6.546 moles of CO2.
Next, we calculate the mass of NaOH and LiOH required. The molar mass of NaOH is 39.997 g/mol, and the molar mass of LiOH is 23.95 g/mol.
Mass of NaOH = Moles of NaOH x Molar mass of NaOH
Mass of NaOH = 6.546 mol x 39.997 g/mol
Mass of NaOH = 261.75 g
Mass of LiOH = Moles of LiOH x Molar mass of LiOH
Mass of LiOH = 6.546 mol x 23.95 g/mol
Mass of LiOH = 156.93 g
Therefore, 261.75 grams of NaOH and 156.93 grams of LiOH are needed to remove 288 grams of CO2 from the air.
c. Based on the calculations, we can see that more NaOH is required compared to LiOH to react with the same amount of CO2. Since NaOH is less expensive per mole, it would be more cost-effective to use NaOH. However, LiOH is used during shuttle missions instead of NaOH due to its less corrosive nature. LiOH is less likely to damage the sensitive equipment and systems in the spacecraft, making it a better choice despite being more expensive per mole.