AP Chemistry: Stoichiometry and Limiting Reagents (the 4-step method)

Stoichiometry shows up in every AP Chem FRQ section. Limiting-reagent problems are the version that costs the most points because students rush the setup. The 4-step method below makes every one of them mechanical.

The 4 steps, every time

  1. Balance the equation. Always. Even if it looks balanced. Even if the question says it is.
  2. Convert each reactant to moles. Grams → moles using molar mass. Volume → moles using density or molarity.
  3. Divide moles by the stoichiometric coefficient. Whichever number is smaller is the limiting reagent.
  4. Use the limiting reagent's moles to compute moles of product, then convert to grams or whatever the question asks.

Worked example

15 g of Al reacts with 50 g of Cl2 to form AlCl3. How much AlCl3 is formed?

Step 1: balance: 2 Al + 3 Cl2 → 2 AlCl3.

Step 2: moles Al = 15 / 27 = 0.556 mol. Moles Cl2 = 50 / 71 = 0.704 mol.

Step 3: Al: 0.556 / 2 = 0.278. Cl2: 0.704 / 3 = 0.235. Cl2 is smaller, so Cl2 is limiting.

Step 4: moles AlCl3 formed = (2/3) × 0.704 = 0.469 mol. Mass = 0.469 × 133.5 = 62.6 g AlCl3.

Common mistakes

  1. Not balancing first. Stoichiometric ratios from an unbalanced equation are meaningless.
  2. Comparing moles directly without dividing by coefficients. 0.556 mol Al is more than 0.704 mol Cl2 only if you ignore that you need 3 Cl2 per 2 Al.
  3. Using the wrong reactant for the product calculation. Once you've identified the limiting reagent, only its moles drive product. Excess reagent moles are ignored.
  4. Forgetting to convert units in the answer. Question asks for grams? Convert from moles. Liters of gas? Use PV = nRT or molar volume.

The percent yield extension

If the question asks for percent yield, the formula is: % yield = (actual yield / theoretical yield) × 100. Theoretical yield is what your stoichiometry calculation gives. Actual yield is provided in the problem. AP Chem regularly includes percent yield as a final step on stoichiometry FRQs.

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