Stoichiometry and Limiting Reagent: Complete Exam Guide

Stoichiometry is the calculation of reactant and product quantities from a balanced chemical equation. The equation's coefficients are mole ratios: 2H2 + O2 → 2H2O says 2 moles of hydrogen react with 1 mole of oxygen to make 2 moles of water. The limiting reagent is the reactant that runs out first and therefore caps how much product can form; the others are in excess.

This matters wherever chemicals are made or used. A factory wasting expensive reactant, a pharmacist compounding a drug, or an engineer sizing an airbag all rely on limiting-reagent thinking. It is also the single most tested calculation skill in school chemistry — moles in, moles out, with the balanced equation as the exchange rate.

Key Concepts

• Stoichiometric ratio: the mole ratio from the balanced equation's coefficients.
• Limiting reagent: the reactant fully consumed first; it decides the maximum product.
• Excess reagent: the reactant left over after the limiting one is gone.
• Theoretical yield: the maximum product mass the limiting reagent allows, assuming perfect reaction.
• Actual yield: what you really obtain in the lab or plant — always equal to or less than theoretical.
• Percentage yield: (actual yield ÷ theoretical yield) × 100; measures reaction efficiency.

Formulas

Percentage yield = (actual yield ÷ theoretical yield) × 100

Finding the limiting reagent (for aA + bB → products):

1. Convert each reactant mass to moles: n = m ÷ M.
2. Divide each mole value by its coefficient: nA/a and nB/b.
3. The smaller result belongs to the limiting reagent.
4. Use the limiting reagent's moles with the equation's mole ratio to find product moles, then convert to mass.

Worked Example

10 g of hydrogen reacts with 60 g of oxygen: 2H2 + O2 → 2H2O. Find the limiting reagent, the mass of water formed, and the leftover reactant.

1. Moles: H2 = 10 ÷ 2 = 5 mol; O2 = 60 ÷ 32 = 1.875 mol.
2. Divide by coefficients: H2: 5 ÷ 2 = 2.5; O2: 1.875 ÷ 1 = 1.875.
3. 1.875 < 2.5, so oxygen is the limiting reagent.
4. Product: 1 mol O2 → 2 mol H2O, so water = 2 × 1.875 = 3.75 mol.
5. Mass of water = 3.75 × 18 = 67.5 g.
6. Leftover H2: used = 2 × 1.875 = 3.75 mol; left = 5 − 3.75 = 1.25 mol = 2.5 g.

Common Misconceptions

• "The reactant with the smallest mass is the limiting reagent." What matters is moles relative to the equation's coefficients, not grams. In the example, hydrogen had the smaller mass (10 g vs 60 g) yet oxygen limited the reaction.
• "All reactants get completely used up." Only the limiting reagent is fully consumed. Excess reagent remains by definition — that is why the worked example ends with 2.5 g of hydrogen untouched.
• "Percentage yield can exceed 100%." It cannot. A yield over 100% means the product is impure or wet, or a weighing error — the extra mass is not extra product.

Real-World Applications

• Fertiliser plants: ammonia synthesis (N2 + 3H2 → 2NH3) is run with carefully chosen excess of the cheaper gas to maximise conversion of the expensive one.
• Pharmaceuticals: costly intermediates are made the limiting reagent deliberately, so none is wasted.
• Airbags: the exact moles of sodium azide are calculated so decomposition yields precisely the gas volume needed — no more, no less.
• Combustion engines: fuel-air ratios are tuned around the stoichiometric point for clean, complete burning.

Stoichiometry turns a balanced equation into a recipe: convert masses to moles, divide by coefficients to find the limiting reagent, and scale everything from it. Remember that the smallest mass is not automatically limiting, and that percentage yield measures real-world efficiency. Master this procedure and quantitative chemistry becomes routine.