Chapter 8 · Quantities in Chemical Reactions
8.6Limiting Reactant, Theoretical Yield, and Percent Yield from Initial Masses
14 min · two checks
Predict
You are given grams of both reactants. How do you find which one limits?
The idea
- Identify the limiting reactant from two given masses.
- Report theoretical yield and percent yield from that choice.
Do not compare grams. Convert both reactants to moles. A clean test: moles available divided by the coefficient. The smaller number is the limiting reactant, and that number is also how many “times” the reaction can run. Alternatively, convert reactant A into moles of product, convert reactant B into moles of the same product, and keep the smaller product amount. Both methods agree if the equation is balanced.
Once you know the limiting reactant, finish the mass-to-mass path to the theoretical yield. Subtracting the used moles of the excess reactant from the starting moles tells you how much excess remains. If an actual yield is given, percent yield is the last step and uses only the theoretical mass you just earned. Label every number. This problem is long only because three familiar problems are stacked.
Keep these
- Convert both reactants to moles before comparing.
- Smaller (moles / coefficient) means limiting.
- Theoretical yield comes only from the limiting reactant.
Worked path
2 Al + 3 Cl₂ → 2 AlCl₃. You combine 2.70 g Al (0.100 mol) with 0.120 mol Cl₂. Which limits, and what mass of AlCl₃ can form? Molar mass AlCl₃ = 133.3 g/mol.
Observe
Al can run 0.100/2 = 0.050 times. Cl₂ can run 0.120/3 = 0.040 times.