Question

The space shuttle environmental control system handles excess (which the astronauts breathe out; it is 4.0% by mass of exhaled air) by reacting it with lithium hydroxide, , pellets to form lithium carbonate, , and water. if there are nine astronauts on board the shuttle, and each exhales 20. l of air per minute, how long could clean air be generated if there were 27,000 g of pellets available for each shuttle mission? assume the density of air is 0.0010 g/ml.

Answer 1

Given:
9 astronauts
volumetric flow rate of air = 20 L / min
mass of Lithium hydroxide = 27000 g 
4% exhaled air
Density of air = 0.0010 g/mL

Solution:

20 L / min (0.0010 g / mL ) (1000 mL / 1 L) = 20.0 g of air / min of each astronaut 

Total mass flow rate of air = 9 * 20.0 g / min = 180.0 g of air / min 

Mass of  CO2 produced = 180 g / min ( 0.04 ) = 7.2 g CO2 / min
The chemical reaction would be expressed as:

CO2 + 2 LiOH  = Li2CO3 + H2O 

moles LiOH = 27000 g of LiOH / 23.95 g / mol
moles LiOH = 1127.35 moles

Dividing the number of moles of LiOH by 2 from the relation in the reaction would allow us to get the number of moles of CO2 that reacts. Then, multiply the resulting value by 44.01 g/mol to get the mass of CO2.

1127.35 mol LiOH ( 1 mol CO2 / 2 mol LiOH ) ( 44.01 g / mol ) = 24807.31 g CO2

Then, divide the mass of CO2 that reacts by the mass flow rate of CO2 which is 0.56 g CO2 per minute,

24807.31 g / 0.56 g /min = 44298.76 minutes