AP Chemistry · Lesson 2

Stoichiometry and limiting reactants

Know Know it

Every stoichiometry problem follows one path: grams of A → moles of A → moles of B → grams of B. Divide by molar mass to get moles, use the mole ratio from the balanced equation to switch substances, then multiply by the molar mass of the new substance. Molarity times volume in liters also gives moles.

When two reactants are given, one of them runs out first. That limiting reactant sets the theoretical yield. To find it, work out how much product each reactant could make, or how much of the other reactant each one needs. The smaller amount of product points to the limiting reactant. The other reactant is in excess.

Percent yield = actual yield ÷ theoretical yield × 100. Keep all digits in your calculator and round only at the end, to the number of significant figures in the least precise measurement.

Worked example Easy

Heating calcium carbonate decomposes it: CaCO₃(s) → CaO(s) + CO₂(g). What mass of CaO forms when 5.00 g of CaCO₃ (100.09 g/mol) decomposes completely? (CaO is 56.08 g/mol.)

  1. 2.20 g
  2. 8.92 g
  3. 1.40 g
  4. 2.80 gAnswer

How to solve it

  1. Grams to moles: 5.00 g ÷ 100.09 g/mol = 0.04996 mol CaCO₃.
  2. Mole ratio 1 : 1, so 0.04996 mol CaO.
  3. Moles to grams: 0.04996 mol × 56.08 g/mol = 2.80 g, which has 3 significant figures like 5.00 g.

Why each choice is right or wrong

  • A. Incorrect. 2.20 g is the mass of CO₂ that forms (0.04996 mol × 44.01 g/mol), not the mass of CaO.
  • B. Incorrect. 8.92 g multiplies 5.00 g by 100.09/56.08, the molar masses upside down; the product mass must be less than 5.00 g here.
  • C. Incorrect. 1.40 g halves the moles of CaO, but the equation has a 1 : 1 ratio between CaCO₃ and CaO.
  • D. Correct. 5.00 g ÷ 100.09 g/mol = 0.04996 mol CaCO₃, which gives 0.04996 mol CaO by the 1 : 1 ratio, and 0.04996 × 56.08 = 2.80 g.

Trap Less mass does not mean limiting

The reactant with the smaller mass, or the smaller number of moles, is not always the limiting reactant. Compare moles through the coefficients: a reactant with a large coefficient can run out first even when more moles of it are present.

Explain Explain it

Task verbs: determine, justify

To justify which reactant is limiting, show one comparison with numbers. For example: "0.50 mol of A would need 1.5 mol of B, but only 1.2 mol of B is present, so B runs out first." Then find every product amount from the limiting reactant only.

The excess left over is the starting amount minus the amount the limiting reactant used up.

Worked example Medium

Aluminum reacts with chlorine gas: 2Al(s) + 3Cl₂(g) → 2AlCl₃(s). A sample of 5.40 g Al (26.98 g/mol) is mixed with 17.7 g Cl₂ (70.90 g/mol). What is the theoretical yield of AlCl₃ (133.33 g/mol)?

  1. 26.7 g
  2. 22.2 gAnswer
  3. 33.3 g
  4. 49.9 g

How to solve it

  1. Moles: 5.40 ÷ 26.98 = 0.2001 mol Al; 17.7 ÷ 70.90 = 0.2496 mol Cl₂.
  2. Al would need 0.2001 × 3/2 = 0.300 mol Cl₂; only 0.2496 mol is there, so Cl₂ is limiting.
  3. AlCl₃ from Cl₂: 0.2496 × 2/3 = 0.1664 mol; × 133.33 g/mol = 22.2 g.

Why each choice is right or wrong

  • A. Incorrect. 26.7 g is the yield if aluminum were limiting; 0.2001 mol Al would need 0.300 mol Cl₂, but only 0.2496 mol is present.
  • B. Correct. Cl₂ is limiting: 0.2496 mol Cl₂ × 2/3 = 0.1664 mol AlCl₃, and 0.1664 × 133.33 = 22.2 g.
  • C. Incorrect. 33.3 g uses a 1 : 1 ratio of Cl₂ to AlCl₃; the equation gives 2 AlCl₃ for every 3 Cl₂.
  • D. Incorrect. 49.9 g multiplies the Cl₂ moles by 3/2 instead of 2/3, so it makes more AlCl₃ than the chlorine can supply.

Justify Your turn

Iron reacts with oxygen gas: 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s). A sealed flask holds 0.200 mol of Fe and 0.250 mol of O₂. Say which reactant is limiting, justify your choice with a calculation that uses the mole ratio, and give the moles of Fe₂O₃ that can form.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Names iron (Fe) as the limiting reactant
  • Justifies the choice with a mole-ratio calculation
  • Gives 0.100 mol of Fe₂O₃

Model answer

Iron is the limiting reactant. By the 4 : 3 ratio, 0.200 mol of Fe needs only 0.200 × 3/4 = 0.150 mol of O₂, and the flask holds 0.250 mol, so the iron runs out first and 0.100 mol of O₂ is left over. The amount of Fe₂O₃ is set by the iron: 0.200 mol Fe × 2/4 = 0.100 mol of Fe₂O₃.

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