IB Chemistry SLTopic 5 — Quantifying Chemical ChangePaper 1 & 2Core skill~11 min read
Limiting and Excess Reactants
Give a reaction two reactants and it will almost never use them up neatly together. One runs out, the reaction stops, and whatever is left of the other just sits there. Everything you calculate afterwards depends on knowing which was which.
📚 What you need to know
The limiting reactant is the one that runs out first and stops the reaction.
The excess reactant is left over when the reaction finishes.
To identify the limiting reactant: find moles of each, then divide by its coefficient. The smallest answer is limiting.
All product calculations use the limiting reactant. The excess one is irrelevant to the yield.
The largest mole count does not mean excess — the coefficients decide.
For gases at the same conditions, you can do the same thing with volumes.
The idea
You have more frames than wheels in absolute numbers — but each bicycle eats two wheels, and that is what settles it.
That analogy contains the whole method. It is not about which reactant you have most of; it is about which one you have most of relative to what the recipe demands. The coefficients are the recipe, and dividing by them is how you compare fairly.
The test
for each reactant, work out moles ÷ coefficient the smallest value is the limiting reactant
Why does dividing work? Because moles ÷ coefficient tells you how many times over the reaction could run on that reactant alone. Whichever supports the fewest runs is the one that stops the show.
🧩 The method
Write and balance the equation.
Convert each reactant to moles — from mass, volume of gas, or concentration × volume.
Divide each by its coefficient.
The smallest result is limiting; everything else is in excess.
Do all further calculations from the limiting reactant, ignoring the others entirely.
Magnesium has more moles than oxygen here, and is still the limiting reactant, because the equation demands two of it for every one O2.
WORKED EXAMPLE
4.00 g of magnesium is burnt in 4.00 g of oxygen. Identify the limiting reactant, calculate the mass of magnesium oxide formed, and the mass of the excess reactant left over. 2Mg(s) + O2(g) → 2MgO(s)
Step 1 — moles of eachn(Mg) = 4.00 ÷ 24.31 = 0.1645 moln(O₂) = 4.00 ÷ 32.00 = 0.1250 molStep 2 — divide by the coefficientsMg: 0.1645 ÷ 2 = 0.0823 O₂: 0.1250 ÷ 1 = 0.1250magnesium is limitingStep 3 — product, from the limiting reactant only2Mg : 2MgO is 1 : 1.n(MgO) = 0.1645 mol → 0.1645 × 40.31 = 6.6326.63 g of MgOStep 4 — oxygen left overO₂ used = 0.1645 ÷ 2 = 0.0823 mol → 0.0823 × 32.00 = 2.63 g4.00 − 2.63 = 1.371.37 g of oxygen unreactedCheck the masses: 4.00 g Mg + 2.63 g O₂ = 6.63 g MgO. Mass is conserved, which is a free check on your answer.
WORKED EXAMPLE
28.0 g of nitrogen is mixed with 8.00 g of hydrogen. Calculate the maximum mass of ammonia obtainable and the mass of the excess reactant remaining. N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 1 — molesn(N₂) = 28.0 ÷ 28.02 = 0.999 moln(H₂) = 8.00 ÷ 2.02 = 3.960 molStep 2 — divide by coefficientsN₂: 0.999 ÷ 1 = 0.999 H₂: 3.960 ÷ 3 = 1.320nitrogen is limitingStep 3 — ammonia, ratio 1 : 2n(NH₃) = 2 × 0.999 = 1.998 mol1.998 × 17.04 = 34.0534.1 g of NH₃Step 4 — hydrogen leftH₂ used = 3 × 0.999 = 2.998 mol(3.960 − 2.998) × 2.02 = 1.941.94 g of hydrogen unreactedFour times as many moles of hydrogen as nitrogen, and hydrogen is still in excess — the reaction only needs three.
The same test works with concentrations and gas volumes. Convert each reactant to moles by whichever route the question hands you, then divide by the coefficient exactly as before. The comparison is always made in moles.
WORKED EXAMPLE
25.0 cm3 of 0.100 mol dm–3 silver nitrate is mixed with 25.0 cm3 of 0.150 mol dm–3 sodium chloride. Calculate the mass of silver chloride precipitated. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
Step 1 — moles of eachn(AgNO₃) = 0.0250 × 0.100 = 2.50 × 10⁻³ moln(NaCl) = 0.0250 × 0.150 = 3.75 × 10⁻³ molStep 2 — both coefficients are 1So the smaller number of moles is simply the limiting one.silver nitrate is limitingStep 3 — mass of precipitateM(AgCl) = 107.87 + 35.45 = 143.322.50 × 10⁻³ × 143.32 = 0.35830.358 g of AgClEqual volumes here, so the concentrations alone decide it — but only because the ratio happens to be 1 : 1.
💡 Exam tip
Divide by the coefficient. Comparing raw moles is the single biggest error in this topic.
The word “excess” in a question is a gift: it tells you the other reactant is limiting, so no test is needed.
Once identified, use the limiting reactant for everything — yield, product mass, gas volume.
For “how much is left over”, find the amount of excess used, then subtract from what you started with.
Check conservation of mass at the end where you can. It catches arithmetic slips instantly.
⚠️ Common mix-up
Assuming the smaller mass is limiting. Mass says nothing until it becomes moles.
Assuming the smaller number of moles is limiting. True only if the coefficients are equal.
Calculating the product from the excess reactant, giving an impossibly large yield.
Forgetting to subtract when asked how much excess remains, and quoting the starting amount.
Ignoring the coefficient when finding how much excess was used. 2Mg per O2 means halving.
Up next: Percentage Yield — the limiting reactant tells you the most you could possibly get, and reality is always somewhat less generous than that.
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