The theoretical yield of H₂O is <u>360</u> g.
We have the masses of two reactants, so this is a <em>limiting reactant</em> problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. <em>Gather all the information</em> 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 <em>moles of each reactant</em>
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 <em>limiting reactant
</em>
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
<em>CO₂ is the limiting reactant </em>because it gives the smaller amount of H₂O.
Step 4. Calculate the <em>theoretical yield</em> of H₂O.
Mass = 20.00 mol H₂O × (18.02 g H₂O /1 mol H₂O) = 360 g H₂O
The theoretical yield is 360 g H₂O.
