The space shuttle environmental control system handles 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 are 7 astronauts on board the shuttle, and each exhales 20. L of air per minute, how long could clean air be generated if there were 28,000 g of LiOH pelle available for each shuttle mission? Assume the density of air is 0.0010 g/mL.
真係好難-.-
Mole Calculation(Hard!!)
2007-02-04 9:58 pm
回答 (2)
2007-02-05 6:51 am
✔ 最佳答案
First of all, the astronauts breath out 7 × 20 × 1000 × 0.001 = 140g of air per minute.Within it, there are 140 × 0.04 = 5.6g of CO2.
Hence the astronauts are breathing out 5.6g of CO2 per minute.
Molar mass of CO2 = 12 + 16 × 2 = 44 g/mol
∴The astronauts are breathing out 5.6/44 = 0.127 moles of CO2 per minute.
Now, from the equation: CO2 + 2LiOH → Li2CO3 + H2O, we can see that 1 mole of CO2 reacts with 2 moles of LiOH completely.
Therefore, moles of LiOH consumed per minute = 0.127 × 2 = 0.254 moles
Now, molar mass of LiOH = 6.9 + 16 + 1 = 23.9 g/mol.
So mass of LiOH consumed per minute = 0.254 × 23.9 = 6.084 g
Finally, the time that the storage of LiOH in the shuttle can last is:
28000/6.084 = 4603 minutes which is about 3.2 days.
參考: My chemical knowledge
2007-02-05 12:20 am
2 LiOH + CO2 -> Li2CO3 + H2O
total CO2 exhausted per minute = 20*1000*0.0010*0.04*7=5.6g
no. of moles of LiOH required per minute = 5.6/(12+16*2) *2 = 0.2545
mass of LiOH required per minute = 0.2545*(6.941)= 1.7665 g
clean air can be generated for 28000/1.7665 = 15850 min = 11 days
total CO2 exhausted per minute = 20*1000*0.0010*0.04*7=5.6g
no. of moles of LiOH required per minute = 5.6/(12+16*2) *2 = 0.2545
mass of LiOH required per minute = 0.2545*(6.941)= 1.7665 g
clean air can be generated for 28000/1.7665 = 15850 min = 11 days
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