Question

The space shuttle environmental control system handled excess CO2 (which the astronauts breathe out; 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 were seven astronauts on board the shuttle, and each exhales 20. L of air per minute, how long could clean air be generated if there were 25,000 g of LiOH pellets available for each shuttle mission? Assume the density of air is 0.0010 g/mL ​

Answer 1

LiOH + CO2 = Li2CO3 + H2O

balance

2LiOH + CO2 = Li2CO3 + H2O

n = 7 astronauts

V = 20 L per min each

find total time for

m = 25000 g to last

First,

find moles of CO2 present:

mol of CO2 = from % of air

mass of air -->volume of air

Volume of air --> 20 L per astronaut--> 7*20 = 140 L per minute

of which

mass = D*V = 140000 * 0.0010 = 140 g of ai

4% is CO2 so --> 4/100*140 = 5.6 g of CO2 per min

mol of CO2 = mass/MW = 5.6/44 = 0.12727 mol of CO2 per min

mol of LiOH present --> mass/MW = 25000/23.95 = 1043.841 mol of LiOH

now...

2 mol of LiOH = 1 mol of CO2

1043.841 /2 = 521.92 mol of CO2

the rate

0.12727 mol of CO2 per min

time = 521.92 /0.12727 = 4100.88 mins --> 4100.88/60 = 68.34 hours --> 2.85 days