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9476 / 2027
The Mole Concept and Stoichiometry

Lesson 3 of 5

Balanced equations and limiting reagents

Use coefficients as mole ratios and identify what runs out first.

A-Level 9476 (2026-2027)

  • I can construct balanced chemical equations.Syllabus

    Syllabus 9476, 6(f). Construct balanced chemical equations.

  • I can calculate reacting amounts in solids, gases and solutions.Syllabus

    Syllabus 9476, 6(g). Calculate reacting amounts in solids, gases and solutions.

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.

Mass -> moles -> stoichiometric ratio -> required quantity
  1. Convert the measured amount

    Use n = m/M, n = cV, or pV = nRT as appropriate.

  2. Apply the balanced coefficients

    If aA reacts with bB, the mole ratio n(A)/a = n(B)/b holds for the amounts actually consumed.

  3. Convert back only at the end

    Find the requested mass, volume, concentration or particle count. Keep extra digits in intermediate calculations.

Worked exampleIdentify 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.

Try it first. Then show the worked stepsHide the worked steps
  1. Balance: 2Al + 3Cl2 → 2AlCl3.
  2. n(Al) = 0.540/27.0 = 0.0200 mol; n(Cl2) = 1.065/71.0 = 0.0150 mol.
  3. Compare amount/coefficient: 0.0200/2 = 0.0100; 0.0150/3 = 0.00500. Chlorine limits the reaction.
  4. n(AlCl3) = (2/3)(0.0150) = 0.0100 mol. M(AlCl3) = 133.5 g mol-1.
Answer

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.

Try it step by step

Work it out in steps: Moles of the first reactant, then Moles of the second reactant, then Moles of product from the limiting reagent, then Maximum mass of product.

  1. Problem 1. 2.30 g of sodium reacts with 2.13 g of chlorine. 2Na + Cl2 -> 2NaCl. Find the maximum mass of NaCl, in g to 3 significant figures. Use Na = 23.0 and Cl = 35.5.
  2. Problem 2. 2.80 g of nitrogen reacts with 0.900 g of hydrogen. N2 + 3H2 -> 2NH3. Find the maximum mass of NH3, in g to 3 significant figures. Use N = 14.0 and H = 1.0.
  3. Problem 3. 1.20 g of magnesium burns in 0.640 g of oxygen. 2Mg + O2 -> 2MgO. Find the maximum mass of MgO. Give each answer to 3 significant figures. Use Mg = 24.0 and O = 16.0.
Show the answer

Problem 1.

Moles of the first reactant: n(Na) = 2.30 / 23.0 = 0.100 mol.

Moles of the second reactant: n(Cl2) = 2.13 / 71.0 = 0.0300 mol.

Moles of product from the limiting reagent: Compare amount / coefficient: Na 0.100/2 = 0.0500; Cl2 0.0300/1 = 0.0300. Chlorine is limiting, so n(NaCl) = 2 x 0.0300 = 0.0600 mol.

Maximum mass of product: Mass = 0.0600 x 58.5 = 3.51 g. 0.0400 mol of Na is left over.

Problem 2.

Moles of the first reactant: n(N2) = 2.80 / 28.0 = 0.100 mol.

Moles of the second reactant: n(H2) = 0.900 / 2.0 = 0.450 mol.

Moles of product from the limiting reagent: Compare amount / coefficient: N2 0.100/1 = 0.100; H2 0.450/3 = 0.150. Nitrogen is limiting, so n(NH3) = 2 x 0.100 = 0.200 mol.

Maximum mass of product: Mass = 0.200 x 17.0 = 3.40 g.

Problem 3.

Moles of the first reactant: n(Mg) = 1.20 / 24.0 = 0.0500 mol.

Moles of the second reactant: n(O2) = 0.640 / 32.0 = 0.0200 mol.

Moles of product from the limiting reagent: Compare amount / coefficient: Mg 0.0500/2 = 0.0250; O2 0.0200/1 = 0.0200. Oxygen is limiting, so n(MgO) = 2 x 0.0200 = 0.0400 mol.

Maximum mass of product: Mass = 0.0400 x 40.0 = 1.60 g.

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.