Potassium superoxide (KO 2 ) is used in spacecrafts to remove carbon dioxide gas

Question

Potassium superoxide (KO 2 ) is used in spacecrafts to remove carbon dioxide gas from the cabin atmosphere and at the same time produce oxygen gas. The CO 2 output per astronaut breathing has been estimated as 1.00 kg per 24 hours period. The potassium superoxide has been proposed to reduce the CO 2 level at a rate of 600 mL (STP) per minute. What fraction of time must the converter operate to keep up with the CO 2 output of one astronaut? The molar mass of KO 2 = 71.10 g/mol and K 2 CO 3 = 138.21 g/mol. 4 KO 2 (s) + 2CO 2 (g) = (sorry can't do the arrow) 2 K 2CO 3 (s) + 3 O 2 (g)

Explanation / Answer

1Kg of CO2 is 1000g/44g/mol = 22.73 moles of CO2

KO2 reduces CO2 @ 600 mL (STP) per minute

PV=nRT

0.6L x 1 atm = n x 0.08206 x 273

n = 0.0268 moles /min

22.73moles/0.0268 moles/min = 848 min which is 848/600= 14.14 h

So the convertor must run for 14.14 h to remove the CO2 produced by one astronaut in 24 h so the fraction of time it has to run is 14.14/24 = 0.59 fraction

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.