Chemistry - Thermochemistry Question with Solution | TestHub
The carbon dioxide exhaled in the breath of astronaut is often removed from the spacecrafts by reaction with lithium hydroxide.
2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)
Assume that each astronaut requires 2100 kcal of energy per day. Further assume that this energy is obtained only from the combustion of glucose into CO2(g) and H2O(l). The standard enthalpies of formation of glucose, CO2 and water are -1280, -395 and -285 kJ/mol, respectively. The minimum mass (in g) of LiOH required per astronaut per day to react completely with all the CO2 produced is
Answer:
Solution:
The combustion enthalpy of glucose, Δ Hcomb(glucose), is calculated as:
[6(-395) + 6(-285)] - (-1280) = -2800 kJ/mol
The energy required is 2100 × 4.184 = 8786 kJ.
Thus, the moles of glucose, nglucose, needed are:
nglucose = 8786/2800 = 3.138 mol
The moles of CO₂, nCO₂, produced are:
nCO₂ = 6 × 3.138 = 18.83 mol
The moles of LiOH needed are 2 × 18.83 = 37.66 mol.
The mass of LiOH needed is 37.66 × 23.94 ≈ 902 g.
Answer: 0902.
