The big idea: A balanced equation is a recipe in moles. The coefficients tell you the mole ratio in which substances react.
Reactants are rarely mixed in the exact ratio, so usually one runs out first — this is the limiting reactant. It controls how much product you can make.
The other reactant is left over: it is in excess.
The theoretical yield is the maximum mass of product, worked out from the limiting reactant.
Three terms to keep straight: - Limiting reactant — the one that runs out first; it sets the amount of product. - Reactant in excess — the one left over after the reaction stops. - Theoretical yield — the maximum product, calculated from the limiting reactant.
The golden rule: never work out a product amount from a reactant that is in excess.
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When one reactant amount is given, the route is always the same: convert the mass to moles, scale by the mole ratio from the equation, then convert back to a mass. Every step starts from the given equation.
- amount of substance (mol)
- mass of the sample (g)
- molar mass (g mol⁻¹), read off the periodic table
Worked example — mass of product from one reactant
Magnesium burns in oxygen: 2Mg + O2 → 2MgO. Calculate the mass of magnesium oxide formed when 4.86 g of magnesium is burned completely. (M(Mg) = 24.31 g mol⁻¹, M(MgO) = 40.31 g mol⁻¹.)
Solution
- Formula first — convert the magnesium mass to moles:
- Use the mole ratio from the equation (2 Mg : 2 MgO, i.e. 1 : 1):
- Convert the product moles back to a mass — keep the unit:
Final answer
m(MgO) = 8.06 g.
Watch the ratio: Here the ratio was 1 : 1, so the moles carried straight across. If the equation had been N2 + 3H2 → 2NH3, then 1 mol of N2 would give 2 mol of NH3 — always read the coefficients.
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When both reactant amounts are given, you must find which one runs out first. Convert each to moles, then divide each amount by its coefficient in the equation — the smallest result is the limiting reactant.
| Step | What you do | Why |
|---|---|---|
| 1 | Write the balanced equation | it gives the mole ratio |
| 2 | Convert each given mass to moles (n = m/M) | the equation works in moles, not grams |
| 3 | Divide each amount by its coefficient | compares them fairly against the ratio |
| 4 | Smallest value = limiting reactant | it runs out first and stops the reaction |
| 5 | Use the limiting reactant's moles + the ratio | to find product moles, then mass |
Worked example — limiting reactant and theoretical yield
Iron reacts with sulfur: Fe + S → FeS. A mixture of 5.60 g of iron and 3.60 g of sulfur is heated. (M(Fe) = 55.85, M(S) = 32.07, M(FeS) = 87.92 g mol⁻¹.) Identify the limiting reactant and calculate the theoretical yield of iron(II) sulfide.
Solution
- Formula first — convert each reactant mass to moles:
- The ratio is 1 : 1 (Fe : S), so divide each by its coefficient (here both are 1) and compare. Iron gives the smaller value, so iron is the limiting reactant (sulfur is in excess).
- Use the limiting amount with the 1 : 1 : 1 ratio to find the moles of FeS:
- Convert to a mass for the theoretical yield:
Final answer
Iron is limiting (sulfur in excess); theoretical yield of FeS = 8.79 g.
Always work from the limiting reactant: If you had used the sulfur (0.112 mol) here, you would have over-estimated the product — there isn't enough iron to react with all of it. The leftover sulfur is the excess.
How this is tested: Reacting masses run right through the course.
- Paper 1A (MCQ): a one-step 'what mass / amount of product forms from this mass of reactant?' or 'which reactant is limiting?' - Paper 2: a part-marks question — 'find the moles of each reactant, deduce the limiting one, then calculate the theoretical yield.'
The trap that loses marks: using the reactant that is in excess, or forgetting to apply the mole ratio when the coefficients are not 1 : 1.
Score it cleanly: (1) Convert every mass to moles first (n = m/M). (2) Divide each amount by its coefficient — smallest = limiting. (3) Work the product only from the limiting reactant, then convert back to a mass with the right significant figures.
IB-style question — magnesium and hydrochloric acid (a)
Magnesium reacts with hydrochloric acid: Mg + 2HCl → MgCl2 + H2. A flask contains 0.486 g of magnesium and 0.0300 mol of HCl. (a) Determine which reactant is limiting. (M(Mg) = 24.31 g mol⁻¹.) [3]
How to score the marks
- Mark 1 — convert the magnesium to moles (the HCl is already in mol):
- Mark 2 — divide each amount by its coefficient (Mg = 1, HCl = 2):
- Mark 3 — smallest value is limiting. HCl gives 0.0150, which is smaller, so HCl is the limiting reactant (magnesium is in excess).
Final answer
HCl is the limiting reactant (magnesium is in excess).
IB-style question — magnesium and hydrochloric acid (b)
(b) Calculate the amount, in moles, of hydrogen gas produced. [2]
How to score the marks
- Mark 1 — use the limiting reactant (HCl) with the mole ratio. From the equation, 2 mol HCl : 1 mol H2, so divide the HCl amount by 2:
- Mark 2 — evaluate:
Final answer
n(H₂) = 0.0150 mol (worked from the limiting HCl, not the excess magnesium).