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. The smallest quotient is the maximum reaction extent in moles for the equation as written. Multiplying it by a product coefficient directly gives that product’s theoretical amount in moles.
For N₂ + 3 H₂ → 2 NH₃ supplied with 2.00 mol N₂ and 3.00 mol H₂, the quotients are 2.00/1 and 3.00/3. The smaller is 1.00 mol of reaction as written, so the maximum product is 2 × 1.00 = 2.00 mol NH₃. This is equivalent to calculating each reactant's possible product amount, because every possible yield is its quotient multiplied by the same product coefficient.
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.
| Substance | Initial amount | Change | Final amount |
|---|---|---|---|
| N₂ | 2.00 mol | −1.00 mol | 1.00 mol |
| H₂ | 3.00 mol | −3.00 mol | 0 mol |
| NH₃ | 0 mol | +2.00 mol | 2.00 mol |
Use the limiting amount to find every change; do not subtract the same numerical amount from different substances. In the aluminum/chlorine example, 30.0 g Cl₂ is 0.4231 mol and consumes (2/3)(0.4231) = 0.2821 mol Al, or 7.61 g. Starting from 10.0 g leaves 2.39 g Al. The limiting reactant has zero left in this complete-reaction model, while the excess amount is nonnegative.
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.
The smallest mass or smallest mole amount need not limit. With 0.600 mol NH₃ and 0.600 mol O₂ in 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O, oxygen is required faster in the recipe and runs out first. If more limiting reactant is added, the identity of the limiting reactant can change; recheck the ratio rather than assuming the same reagent limits forever. Equilibrium and incomplete conversion can leave some of both reactants in an actual flask; the theoretical yield here assumes completion.
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.
Sometimes neither reactant is in excess. If nᵢ divided by its equation coefficient is the same for both reactants, they are supplied in the exact reaction ratio and both can run out together.
For 2 H₂ + O₂ → 2 H₂O, 4.0 mol H₂ and 2.0 mol O₂ give 4.0/2 = 2.0 and 2.0/1 = 2.0. There is no unique limiting reactant and no leftover reagent. If the quotients differ, the smaller quotient limits.
This comparison works before any product calculation and prevents forcing a 'limiting' label onto a perfectly matched mixture.
The 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.
For 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O, 0.600 mol of each reactant is supplied. Find the limiting reactant, NO produced and NH₃ left.
Try it first: Compare 0.600/4 with 0.600/5.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Calling the reactant with the fewest moles the limiting one. | Coefficients set the consumption rate, so a reactant with more moles can still run out first, as H₂ does in the ammonia example. | Divide each reactant's moles by its coefficient, or compute the product each could make. |
| Comparing masses in grams to decide which reactant limits. | Molar masses differ, so 30 g of a light reactant can be far more moles than 30 g of a heavy one. | Convert every reactant to moles before comparing anything. |
| Calculating the theoretical yield from the excess reactant. | That reactant is still present when the reaction stops, so it overstates the product. | Always base the yield on the reactant that gave the smaller product amount. |
| Subtracting moles of limiting reactant directly from moles of excess reactant. | The two are consumed in the coefficient ratio, not one for one. | Convert the limiting amount into moles of the excess reactant consumed before subtracting. |
| Assuming adding more excess reactant produces more product. | The limiting reactant has already been fully consumed, so nothing further can react. | Ask what would still be available to react, then answer. |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
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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.
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