# Answer to Question #1091 in Inorganic Chemistry for Thomasina

Question #1091

One method of removing CO[sub]2[/sub] gas from a spacecraft is to allow the CO[sub]2[/sub] to react with LiOH.
How many liters of CO[sub]2[/sub] at 25.9 degrees Celcius and 751 Torr can be removed per kilogram of LiOH consumed?

Expert's answer

CO

We have 1kg LiOH : 1000g/(7+16+1) g/mol = 1000/24 = 41.67 mol.

Thus M(CO

m(CO

Use the equation

pV = ν R T

p = 751 torr = 100 125.099 Pa, T = 25.9 C = 298.9 K

V = ν R T / p = 83.33 * 8.31 *298.9 /100125.099 = 2.07 m

_{2}+ 2LiOH = Li_{2}CO_{3}+ H_{2}OWe have 1kg LiOH : 1000g/(7+16+1) g/mol = 1000/24 = 41.67 mol.

Thus M(CO

_{2}) = 2 M(LiOH) = 83.33 molm(CO

_{2}) = 83.33 mol * (12 + 2*16) g/mol = 4000 g = 4 kg.Use the equation

pV = ν R T

p = 751 torr = 100 125.099 Pa, T = 25.9 C = 298.9 K

V = ν R T / p = 83.33 * 8.31 *298.9 /100125.099 = 2.07 m

^{3}= 2067.2 Liters
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