What you'll be able to do: Identify which reactant runs out first, calculate the maximum product, and find how much of the excess reactant is left over.
Best after: Mole Ratios and Mass-to-Mass Stoichiometry
Real reactions rarely start with exactly matching amounts. One reactant runs out first and stops the reaction, and everything that is left of the other simply sits there unreacted. Finding that limiting reactant is the step that turns a textbook ratio into a prediction about a real flask.
These are recommended, not required. You can start this lesson at any time.
With 10 slices of bread and 3 slices of cheese you can make 3 sandwiches, and 4 slices of bread are left over. The cheese limits. Chemistry works the same way, with one important difference: the recipe ratio comes from the coefficients, so it is rarely 1:1. Always check the ratio before you compare amounts.
The most reliable method has three steps. 1. Convert each reactant's mass to moles. 2. For each reactant separately, use its mole ratio to calculate how much of one chosen product it could make. 3. The reactant that yields the smaller amount is the limiting reactant, and that smaller amount is the theoretical yield. It works for any number of reactants and any coefficients.
nproduct from A = nA x (coeffproduct / coeffA)
An alternative is to divide each reactant's moles by its own coefficient. The smallest quotient identifies the limiting reactant. This is faster, but it only tells you which reactant limits, you still have to run one full calculation for the yield.
nA / coeffA versus nB / coeffB
Excess left over = starting moles of excess minus moles consumed, where the consumed amount is found from the limiting reactant using the mole ratio between the two reactants. Convert back to grams at the end if the question asks for mass. A common check: the leftover can never exceed what you started with, and it can never be negative.
nexcess remaining = nexcess initial - nlimiting x (coeffexcess / coefflimiting)
Saying a reactant is in excess is a statement about the amounts supplied, not about the substance itself. The same reactant can be limiting in one experiment and in excess in another. Adding more of the excess reactant does not increase the yield, because the limiting reactant is already gone.
If a calculated theoretical yield is smaller than a measured actual yield, something is wrong: usually an unbalanced equation, the wrong limiting reactant, or an impure product that still contains solvent or water. Treat an impossible percent yield above 100 % as a diagnostic signal rather than a rounding issue. This kind of reasoning appears in the next lesson.
Next: Percent YieldThe reactant that is fully consumed first, capping how much product can form.
Any reactant that remains after the limiting reactant has been used up.
The maximum amount of product possible, calculated from the limiting reactant assuming complete reaction.
Comparing raw moles of reactants is only valid when their coefficients are equal. Comparing the product each could make is always valid.
Once the limiting reactant is gone the reaction stops, no matter how much of the other reactant remains.
nP(from A) = (mA / MA) × (coeffP / coeffA)
nleft = nexcess initial - nlimiting x (coeffexcess / coefflimiting)
N₂ + 3 H₂ → 2 NH₃. You have 2.0 mol N₂ and 3.0 mol H₂. Which reactant is limiting, and how much NH₃ forms?
Try it first: Predict which is limiting before calculating. Was the reactant with fewer moles the one that limited?
0 of 4 steps revealed.
2 Al + 3 Cl₂ → 2 AlCl₃. Starting with 10.0 g Al and 30.0 g Cl₂, what mass of AlCl₃ forms?
0 of 4 steps revealed.
Why it's wrong: Coefficients set the consumption rate, so a reactant with more moles can still run out first, as H₂ does in the ammonia example.
Check instead: Divide each reactant's moles by its coefficient, or compute the product each could make.
Why it's wrong: Molar masses differ, so 30 g of a light reactant can be far more moles than 30 g of a heavy one.
Check instead: Convert every reactant to moles before comparing anything.
Why it's wrong: That reactant is still present when the reaction stops, so it overstates the product.
Check instead: Always base the yield on the reactant that gave the smaller product amount.
Why it's wrong: The two are consumed in the coefficient ratio, not one for one.
Check instead: Convert the limiting amount into moles of the excess reactant consumed before subtracting.
Why it's wrong: The limiting reactant has already been fully consumed, so nothing further can react.
Check instead: Ask what would still be available to react, then answer.
No practice questions are available for this topic yet. You can still practice the whole unit.
Convert every reactant to moles, use the mole ratio to find how much product each one could make on its own, and take the smallest result. That reactant is limiting, its value is the theoretical yield, and the leftover excess is found by subtracting what actually reacted.
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License