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) + H2??O(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).

Answer 1

Given :

Calorie requirement per astrounaut per day2150 Cal

Calorie requirement by 3 astrounaut in 7 days(2150 Cal/day)*(7days)*345150 Cal45150 Cal*(4.184J/Cal)188907.6J

Now from the combustion equation for glucose,

C6H12O6 + 6O2 --->6CO2 +6H2O

Enthalpy of this rxn ,delta H (glucose)-1273.3 KJ/mol(-ve sign indicates heat released)

But the metabolism is only 66% efficient,

Heat released per mol of glucose combustion0.66*(1273.3 KJ/mol)840.378 KJ/mol

So, mol of glucose required to produce 188907.6 J 188907.6J/(840.378KJ/mol)

188907.6J/(840.378 *1000J/mol)0.225mol

mol of CO2 : mol of glucose6:1

So, mol of CO2 produced6*mol of glucose6*0.225 mol1.35 mol

mol of LiOH required per mol CO2 2 mol

mol of LiOH :mol of CO2 2:1

So mol of LiOH 2*1.35 mol2.7 mol

mass of LiOH requiredmol *molar mass(2.7 mol)*(23.95 g/mol)64.665 g0.0647 Kg