Topic 3 of 5
Balanced equations and limiting reagents
Use coefficients as mole ratios and identify what runs out first.
A-Level 9476 (2026-2027)
The balanced equation is the conversion between substances
Balance atoms and charge; change coefficients, never formula subscripts.
A balanced equation conserves each element and total charge. For an ionic equation, remove unchanged spectator ions, then check atoms and charge again. For example, CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l) has zero net charge on both sides. Replacing CO3 with CO2 to balance oxygen would change the reacting species and is invalid.
- Convert the measured amount
Use n = m/M, n = cV, or pV = nRT as appropriate.
- Apply the balanced coefficients
If aA reacts with bB, the mole ratio n(A)/a = n(B)/b holds for the amounts actually consumed.
- Convert back only at the end
Find the requested mass, volume, concentration or particle count. Keep extra digits in intermediate calculations.
Worked example
Identify the limiting reagent before calculating yield
0.540 g Al reacts with 1.065 g Cl2. Find the maximum mass of AlCl3. Use Al = 27.0 and Cl = 35.5.
- Balance: 2Al + 3Cl2 → 2AlCl3.
- n(Al) = 0.540/27.0 = 0.0200 mol; n(Cl2) = 1.065/71.0 = 0.0150 mol.
- Compare amount/coefficient: 0.0200/2 = 0.0100; 0.0150/3 = 0.00500. Chlorine limits the reaction.
- n(AlCl3) = (2/3)(0.0150) = 0.0100 mol. M(AlCl3) = 133.5 g mol-1.
Maximum mass = 1.34 g to three significant figures. Only 0.0100 mol Al reacts, leaving 0.270 g Al. Theoretical yield assumes the specified reaction goes to completion without side reactions.
If the isolated dry product is less than this maximum, percentage yield = actual yield / theoretical yield × 100%. Impure or wet product can give an apparent yield above 100%; this is evidence about the measurement, not extra creation of matter. Distinguish product yield from the percentage purity of a starting sample.