Question

Carbon dioxide scrubbers containing lithium hydroxide continually work to remove exhaled CO2(g) from the breath of...

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).

Homework Answers

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

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