Question

On the space shuttle, lithium hydroxide is produced from lithium oxide and water. The lithium hydroxide then reacts with the exhaled carbon dioxide to remove it from the air supply. The process occurs based on the following reactions: Li2O (s) + H2O (g) ĺ2LiOH (s) LiOH (s) + CO2 (g) ĺLiHCO3 (s) If you have 2.6 g of Li2O and 1.3 g of H2O, how many grams of CO2 can you remove from the air? Report your answer to two significant figures. Do NOT include units in your answer. __1__ g CO2

Answer 1

6,2g of CO₂

Step-by-step explanation:

Based on the reactions:

Li₂O(s) + H₂O(g) → 2LiOH(s)

LiOH(s) + CO₂(g) → LiHCO₃(s)

2,6g of Li₂O and 1,3g of H₂O are:

2,6g × ( 1mol / 29,88g) = 0,087 moles

1.3g × ( 1mol / 18,01g) = 0,072 moles

That means limiting reactant is H₂O. The moles produced of LiOH are:

0,072 moles of H₂O × ( 2mol LiOH / 1mol H₂O) = 0,14 moles of LiOH

Thus, the maximum CO₂ that can react are 0,14 moles of CO₂, in grams

0,14 moles CO₂ × (44,01g / 1mol) = 6,2g of CO₂

I hope it helps!