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_{2} + 2LiOH = Li_{2}CO_{3} + H_{2}O

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

Thus M(CO_{2}) = 2 M(LiOH) = 83.33 mol

m(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

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

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