Potassium superoxide (KO2) is used in spacecrafts to remove carbon dioxide gas f

Question

Potassium superoxide (KO2) is used in spacecrafts to remove carbon dioxide gas from the cabin atmosphere and at the same time produce oxygen gas. The CO2 output per astronaut breathing has been estimated as 1.00 kg per 24 hours period. The potassium superoxide has been proposed to reduce the CO2 level at a rate of 600 mL (STP) per minute. What fraction of time must the converter operate to keep up with the CO2 output of one astronaut? The molar mass of KO2 = 71.10 g/mol and K2CO3 = 138.21 g/mol.

4 KO (s) 2 CO (g) ---> 2 K CO (s) 3 O (g)

Explanation / Answer

4 KO2 + 2CO2 --> 2K2CO3 + 3O2

Consider 24 hour period

Moles of CO2 generated, Mg = 1 * 1000 / 44 = 22.73 moles

CO2 reduction rate = 600 mL/ min

Using ideal gas equation to estimate moles of CO2 reduction rate at STP, we get

PV = nRT

1 * 600 /1000 = n * 0.0821 * 273.15

n = 0.027 moles/min

During 24 hour period

Moles of CO2 reduced, Mr = n * 60 * 24 = 38.53 moles

Fraction of time the converter must operate = Mg/Mr = 22.73 / 38.53 = 0.59

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