The theoretical yield of Hâ‚‚O is 360 g. Â
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. Â
Step 1. Gather all the information in one place with molar masses above the formulas and everything else below them. Â
M_r: Â Â Â Â Â 44.01 Â Â Â Â Â 23.95 Â Â Â Â Â Â Â Â Â 18.02
         CO₂ +    2LiOH → Li₂CO₃ + H₂O
Mass/g: 8.80 × 10²  1.000 × 10³
Step 2. Calculate the moles of each reactant Â
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 Hâ‚‚O we can obtain from each reactant. Â
From CO₂ : Moles of H₂O = 20.00 mol CO₂ × (1 mol H₂O /1 mol CO₂)
= 20.00 mol Hâ‚‚O
From LiOH: Moles of H₂O = 41.75 mol LiOH × (1 mol H₂O /2 mol LiOH)
= 20.88 mol Hâ‚‚O
COâ‚‚ is the limiting reactant because it gives the smaller amount of Hâ‚‚O.
Step 4. Calculate the theoretical yield of Hâ‚‚O.
Mass = 20.00 mol H₂O × (18.02 g H₂O /1 mol H₂O) = 360 g H₂O
The theoretical yield is 360 g Hâ‚‚O.
