Carbon dioxide scrubbers containing lithium hydroxide continually work to remove
ID: 542770 • Letter: C
Question
Carbon dioxide scrubbers containing lithium hydroxide continually work to remove exhaled CO2(g) from the breath of astronauts on board the international space station:
2LiOH(s) + CO2(g) Li2CO3(s) + H2O(l)
If there are three astronauts aboard the space station, and each astronaut requires 2150 Cal from food per day, what mass (kg) of LiOH is required to remove 100% of the CO2 produced by the three astronauts in seven days? (Assume that the astronauts' energy requirements are met exclusively by the combustion of glucose, H of glucose = -1273.3 kJ/mol. The metabolism of glucose is only 66% efficient).
Explanation / Answer
if food calories are considered then
1 Food calorie =1 Kcal (in actual)
The combustion equation for glucose is
C6H12O6 + 6O2 ---> 6CO2 + 6H2O
The heat released by one mole of glucose = -1273.3 KJ /mol
The actual heat released in the metabolism = 66% X 1273.3 = -840.378 KJ / mole
The energy required by each astronaut /per day = 2150 food calories = 2150 Kcal = 2150 X 4.184 KJ = 8995.6 KJ
The energy required by three astronaut / day = 3 X 8995.6 = 26986.8 KJ
The energy required by three astronaut for seven days = 7 X 26983.8 = 188907.6 KJ
The moles of glucose required = 188907.6 / 840.378 KJ = 224.79 moles of glucose
these moles of glucose will give = 6 X 224.79 moles of CO2 = 1348.74 moles of CO2
For each mole of CO2 we need 2 moles of LiOH
So for 1348.74 moles of CO2 = 2 X 1348.74 Moles of KOH will be required = 2697.48 moles of LiOH
The mass of LiOH required = Moles X molecular weight of LiOH = 2679.48 X 24 = 64739.52 grams = 64.739 Kg