The space shuttle environment control system handles excess CO2 (it is 4.0% by mass of exhaled air) by reacting it with lithium hydroxide, LiOH, pellets to form lithium carbonate,Li2CO3, and water. If there are 7 astronauts on board and each exhales 20. L of air per minute, how long could clean air be generated if there were 25,000g of LiOH available. Assume the density of air to be 0.0010 g/ml
2 answers
Hydroxonitrat aci
first you balance the equation:
CO2+2Li(OH)-->Li2(CO3)+H2O
so it's 20L/min of air per passenger so that's 20*7 gives you 140L/min and then you find 4% of that which is 5.6L/min CO2. then you make a stoich table:
5.6L CO2/min * 1000mL/1 L * 0.0010g CO2/1 ml * 1 mol CO2/44.01g
* 2 mol LiOH/1 mol CO2 * 23.95 g/1 mol LiOH * 1/25000g
so then you multiply all the top and divide by all of the bottom and get: 2.4e-4 min^-1 (minutes are on the bottom) so just inverse that and you get 4108 minutes and that should be the answer.
This is about 2.85 days or about 2 days 20 hours 18 minutes and 21 seconds.
CO2+2Li(OH)-->Li2(CO3)+H2O
so it's 20L/min of air per passenger so that's 20*7 gives you 140L/min and then you find 4% of that which is 5.6L/min CO2. then you make a stoich table:
5.6L CO2/min * 1000mL/1 L * 0.0010g CO2/1 ml * 1 mol CO2/44.01g
* 2 mol LiOH/1 mol CO2 * 23.95 g/1 mol LiOH * 1/25000g
so then you multiply all the top and divide by all of the bottom and get: 2.4e-4 min^-1 (minutes are on the bottom) so just inverse that and you get 4108 minutes and that should be the answer.
This is about 2.85 days or about 2 days 20 hours 18 minutes and 21 seconds.