Advanced stoichiometry & limiting reagents
The idea
Real reactions rarely come with reagents in the exact balanced-equation ratio, so one reactant runs out first — the limiting reagent — and it alone fixes how much product can form. You already convert grams to moles and use mole ratios; the advanced layer is the full pipeline: identify the limiting reagent by comparing supply to stoichiometric demand, compute the theoretical yield from it, then report percent yield = (actual/theoretical) × 100 to quantify how well the real reaction performed.
The reliable test for the limiting reagent is a demand calculation: pick one reactant, compute how much partner it requires through the mole ratio, and compare with what is actually present. Equivalently, divide each reactant's moles by its coefficient and the smallest quotient loses. The trap is comparing raw grams or even raw moles — a reactant can be present in greater mass and greater mole count and still be limiting if the equation demands it in a higher ratio.
Percent yield below 100 is normal (side reactions, incomplete conversion, transfer losses); a value above 100 signals an error or a wet, impure product, not a bonus. Always sanity-check that your theoretical yield respects conservation of mass.
Worked example
Aluminum burns in chlorine by 2 Al + 3 Cl₂ → 2 AlCl₃. A run combines 10.0 g of Al (27.0 g/mol) with 30.0 g of Cl₂ (70.9 g/mol) and isolates 31.2 g of AlCl₃ (133.3 g/mol). Identify the limiting reagent and find the theoretical and percent yields.
- Convert both reactants to moles: 10.0/27.0 = 0.370 mol Al and 30.0/70.9 = 0.423 mol Cl₂.
- Run the demand test: consuming all 0.370 mol Al would require 0.370 × (3 mol Cl₂/2 mol Al) = 0.556 mol Cl₂, but only 0.423 mol is on hand — chlorine is the limiting reagent, even though it outweighs and outnumbers the aluminum.
- Build the theoretical yield from the limiting reagent: 0.423 mol Cl₂ × (2 mol AlCl₃/3 mol Cl₂) = 0.282 mol AlCl₃, and 0.282 × 133.3 = 37.6 g possible at best.
- Compare with reality: percent yield = 31.2/37.6 × 100 = 83.0%.
- Sanity-check: the actual yield sits below the theoretical maximum, as it must, and the leftover Al (0.370 − 0.423 × 2/3 = 0.088 mol) confirms aluminum was in excess.
Answer. Cl₂ is the limiting reagent; the theoretical yield is 37.6 g of AlCl₃ and the percent yield is 83.0%.
Check your understanding
- Why can a reactant present in both greater mass and greater mole count still be the limiting reagent?
- How would you redesign the reactant amounts above so that neither reagent is left over, and when would a chemist deliberately avoid that?
- What physical causes push percent yield below 100, and what should you suspect when a student reports 104%?
- How does the divide-by-coefficient shortcut for finding the limiting reagent follow from the demand-comparison method?
Build the foundations first
Advanced stoichiometry & limiting reagents builds on these concepts. If any feel shaky, start there.