The theoretical yield of H₂O is 360 g.
We have the masses of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. Gather all the information in one place with molar masses above the formulas and everything else below them.
M_r: 44.01 23.95 18.02
CO₂ + 2LiOH → Li₂CO₃ + H₂O
Mass/g: 8.80 × 10² 1.000 × 10³
Step 2. Calculate the moles of each reactant
Moles of CO₂ = 8.80 × 10² g CO₂ × (1 mol CO₂ /44.01 g CO₂) = 20.00 mol CO₂
Moles of LiOH = 1.000 × 10³ g LiOH × (1 mol LiOH /23.95 g LiOH)
= 41.75 mol LiOH
Step 3. Identify the limiting reactant
Calculate the moles of H₂O we can obtain from each reactant.
From CO₂ : Moles of H₂O = 20.00 mol CO₂ × (1 mol H₂O /1 mol CO₂)
= 20.00 mol H₂O
From LiOH: Moles of H₂O = 41.75 mol LiOH × (1 mol H₂O /2 mol LiOH)
= 20.88 mol H₂O
CO₂ is the limiting reactant because it gives the smaller amount of H₂O.
Step 4. Calculate the theoretical yield of H₂O.
Mass = 20.00 mol H₂O × (18.02 g H₂O /1 mol H₂O) = 360 g H₂O
The theoretical yield is 360 g H₂O.
