IB Chemistry HLTopic 5 — How Much? Quantifying Chemical ChangePaper 1 & 2Core skill~10 min read
Limiting and Excess Reactants
Mix reactants in the exact ratio from the equation and everything is used up neatly. Real reactions are almost never like that. One reactant runs out first, and from that moment nothing else can happen — no matter how much of the other one is sitting there.
📚 What you need to know
The limiting reactant runs out first and controls how much product you can make.
The excess reactant is left over when the reaction stops.
To find the limiting one: divide the moles of each reactant by its coefficient. The smallest answer is limiting.
You cannot just compare masses, or even moles, without using the coefficients.
Every calculation afterwards — product mass, gas volume, yield — uses the limiting reactant only.
For gases at the same conditions you can divide volumes by the coefficients instead of moles.
Words like “in excess” in a question are telling you which reactant to ignore.
The test that always works
Students often compare moles and stop there. That fails as soon as the ratio is not 1 : 1. If a reaction needs three moles of hydrogen for every mole of nitrogen, then having “more moles of hydrogen” proves nothing — you need three times more.
Dividing by the coefficient fixes this. It converts each reactant into “how many times could I run this reaction with what I have”, and the reactant that gives the smallest number is the one that stops you.
The limiting reactant test
for each reactant, work out n ÷ coefficient the smallest value is the limiting reactant
There are more moles of zinc than acid here, but that is not why the acid limits. It limits because the reaction needs two acid particles for every zinc atom.
What “excess” looks like in the flask
It helps to stop thinking in grams for a moment and count molecules instead.
Each reaction event uses one N2 and three H2. With six H2 you can run it twice, which makes four NH3 and leaves two N2 untouched.
This is why “there are more moles of nitrogen” is not an answer. What matters is how many complete sets of reactants you can assemble, and that is exactly what dividing by the coefficient measures.
Limiting or excess: what each one is for
Limiting reactant
Excess reactant
What happens to it
Completely used up
Some is left at the end
How you spot it
Smallest n ÷ coefficient
Any larger value
Use it to find
Product mass, volume, yield
How much is left over
In the exam wording
“reacts completely”
“in excess”, “excess acid added”
🧩 The method, and how to find the leftovers
Balance the equation. You need the coefficients.
Work out the moles of each reactant from mass, concentration or volume.
Divide each by its coefficient. The smallest value is limiting.
Use the limiting reactant’s moles and the mole ratio to find whatever the question asks for.
For the leftover: work out how much of the excess reactant was used, then subtract from what you started with.
Worked examples
WORKED EXAMPLE
5.00 g of zinc is added to 100 cm3 of 0.500 mol dm−3 HCl. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g). Which reactant is limiting, and what volume of hydrogen is made at STP?
Step 1: moles of each reactantn(Zn) = 5.00 ÷ 65.38 = 0.0765 mol n(HCl) = 0.100 × 0.500 = 0.0500 molStep 2: divide by the coefficientsZn: 0.0765 ÷ 1 = 0.0765 HCl: 0.0500 ÷ 2 = 0.0250 ← smallestStep 3: HCl is limiting, so use its moles
2 HCl gives 1 H2, so n(H₂) = 0.0500 ÷ 2 = 0.0250 molStep 4: moles to gas volumeV = 0.0250 × 22.7 = 0.5675 dm³HCl is limiting; 0.568 dm³ (568 cm³) of H₂Use the zinc by mistake and you get 1.74 dm³ — three times too much, because there simply is not enough acid to dissolve all that zinc.
WORKED EXAMPLE
14.0 g of nitrogen is mixed with 6.00 g of hydrogen: N2(g) + 3H2(g) → 2NH3(g). Find the maximum mass of ammonia, and the mass of the excess reactant left over.
Step 1: moles of eachn(N₂) = 14.0 ÷ 28.02 = 0.500 mol n(H₂) = 6.00 ÷ 2.02 = 2.970 molStep 2: divide by coefficientsN₂: 0.500 ÷ 1 = 0.500 ← smallest H₂: 2.970 ÷ 3 = 0.990Step 3: nitrogen is limiting, so find the ammonia
1 N2 gives 2 NH3: n(NH₃) = 1.00 mol, m = 1.00 × 17.04 = 17.0 gStep 4: how much hydrogen was used?3 × 0.500 = 1.50 mol used, so 2.970 − 1.50 = 1.47 mol leftm = 1.47 × 2.02 = 2.97 g17.0 g of NH₃, with 2.97 g of H₂ left overThere were nearly six times more moles of hydrogen, and it still was not the limiting one. Only the divided values decide.
WORKED EXAMPLE
30 cm3 of methane is sparked with 100 cm3 of oxygen: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). What gases remain, and what is the total volume?
Step 1: for gases, use volumes directlyCH₄: 30 ÷ 1 = 30 O₂: 100 ÷ 2 = 50Step 2: methane is limitingO₂ used = 2 × 30 = 60 cm³, so 40 cm³ is leftStep 3: product volumesCO₂ = 30 cm³; water is (l) so counts as zeroStep 4: total gas remaining40 + 30 = 70 cm³70 cm³: 40 cm³ unreacted O₂ and 30 cm³ CO₂No moles, no molar masses. With gases at the same conditions, volumes behave exactly like moles, so the same test works on them.
💡 Exam tip
Always divide by the coefficient. Write both divided values down side by side so the comparison is visible.
Say which reactant is limiting in words before you calculate anything else. That statement is usually a mark.
Read the question for free information. “Excess acid” or “excess oxygen” means you do not have to do the test at all.
Never use the excess reactant for a product calculation, however tempting its numbers look.
For leftovers, subtract what was used from what you started with — do not try to do it in one step.
Convert leftovers back into grams or cm3 if the question asks for mass or volume.
⚠ Common mix-up
Comparing masses. 14 g of one substance and 6 g of another tells you nothing about which runs out.
Comparing moles without the coefficients. The most common error, and it gives the wrong reactant whenever the ratio is not 1 : 1.
Assuming the smaller mass is limiting. Hydrogen has a tiny molar mass, so a small mass can be a lot of moles.
Using the limiting reactant to find the leftover. The leftover belongs to the excess reactant.
Forgetting the excess in a gas volume total. Unreacted gas is still gas.
Thinking the excess reactant does nothing. It reacts too — just not all of it.
The limiting reactant gives you the absolute maximum product you could get. In a real lab you never quite reach it, and the gap between the two has its own name. Up next: Percentage Yield.
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