Final answer:
To find the limiting reactant, the available masses of lithium and water are converted to moles. Based on stoichiometry, Li2O, formed from the available lithium, is the limiting reactant because less moles of it can be produced than the moles of water available.
Step-by-step explanation:
To determine the limiting reactant between lithium oxide (Li₂O) and water (H₂O) when used to remove water from the air supply aboard a space shuttle, we must consider the stoichiometry of the reaction:
Li₂O(s) + H₂O(g) → 2LiOH(s)
First, we need to convert the available masses of lithium (Li) and water (H₂O) into moles. For lithium, the atomic mass is approximately 6.94 g/mol, so 69 kg (or 69000 g) of lithium divided by 6.94 g/mol will equal the moles of lithium present. Since Li₂O contains two lithium atoms, the amount of Li₂O that can be formed is half of the moles of lithium available.
For water, the molecular weight is approximately 18.02 g/mol, so 80 kg (or 80000 g) of water divided by 18.02 g/mol gives us the moles of water present.
Now, we can use the stoichiometry of the reaction to find out which reactant is limiting. The stoichiometry tells us that 1 mole of Li₂O reacts with 1 mole of H₂O to produce 2 moles of LiOH. We compare the moles of Li₂O that can be produced from the available lithium to the moles of H₂O available to determine which one is the limiting reactant.
Based on the stoichiometry, if 69 kg of Li is available for the reaction, this means that we have less moles of Li₂O than required to react with the entire 80 kg of H₂O, indicating that Li₂O is the limiting reactant since it will be consumed first during the reaction.