Limiting Reactants Study Pack

Stoichiometric Ratios and Real-World Quantities

• •A balanced equation's coefficients express the exact mole ratio in which reactants are consumed — for example, N₂ + 3 H₂ → 2 NH₃ requires three moles of H₂ for every one mole of N₂.
• •In a laboratory or industrial setting, reactants are rarely added in perfectly stoichiometric amounts, so one will run out first.

The Limiting Reactant Defined

• •The reactant that is completely consumed first is called the limiting reactant (or limiting reagent); it controls, or 'limits,' the maximum amount of product that can form.
• •All other reactants present in greater-than-required amounts are called excess reactants; they are only partially consumed and some quantity of each remains when the reaction ends.

Everyday Analogy for the Concept

• •Consider assembling sandwiches that each require 2 slices of bread and 1 slice of cheese: if you have 10 slices of bread but only 3 slices of cheese, the cheese is limiting because it runs out after 3 sandwiches, leaving 4 bread slices unused.
• •This mirrors chemical stoichiometry exactly — the ingredient present in the smallest ratio relative to what the recipe demands is the limiting component.

• •Step 1 — Convert All Reactant Masses to Moles
• •Divide the given mass of each reactant by its molar mass (g/mol) to obtain the number of moles available.
• •For example, if 28 g of N₂ (molar mass 28.02 g/mol) and 9 g of H₂ (molar mass 2.02 g/mol) are provided, you have approximately 1.0 mol N₂ and 4.5 mol H₂.
• •Method 1 — Compare Mole Ratios Directly
• •Divide the moles of each reactant by its stoichiometric coefficient from the balanced equation.
• •The reactant that gives the smallest quotient is the limiting reactant — it is most deficient relative to what the reaction demands.
• •Using the N₂/H₂ example: 1.0 mol N₂ ÷ 1 = 1.0; 4.5 mol H₂ ÷ 3 = 1.5. N₂ gives the smaller quotient, so N₂ is the limiting reactant.
• •Method 2 — Calculate Product Yield from Each Reactant Separately
• •Assume each reactant is limiting in turn and calculate how many moles of product each would generate if fully consumed.
• •The reactant that produces the smaller amount of product is the limiting reactant, because that smaller quantity is all the reaction can actually make.
• •Both methods yield the same answer; Method 1 is faster for simple reactions, while Method 2 is more intuitive for students new to the concept.

Theoretical Yield from the Limiting Reactant

• •The theoretical yield is the maximum mass of product that the reaction could produce if the limiting reactant were converted completely and no product were lost.
• •To calculate it: use the moles of the limiting reactant, apply the mole-to-mole ratio between the limiting reactant and the product from the balanced equation, then convert product moles to grams using the product's molar mass.
• •In the N₂ + 3 H₂ → 2 NH₃ example with 1.0 mol N₂ limiting: 1.0 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 2.0 mol NH₃; 2.0 mol × 17.03 g/mol ≈ 34.1 g NH₃ theoretical yield.

Amount of Excess Reactant Consumed and Remaining

• •Calculate how many moles of the excess reactant are actually consumed by the limiting reactant using the balanced equation's mole ratio.
• •Subtract consumed moles from the initial moles of the excess reactant to find the amount remaining after the reaction.
• •In the example, 1.0 mol N₂ consumes 3.0 mol H₂ (from the 1:3 ratio), leaving 4.5 − 3.0 = 1.5 mol H₂ unreacted.
• •This leftover amount can be converted back to grams if a mass answer is needed: 1.5 mol H₂ × 2.02 g/mol ≈ 3.0 g H₂ remaining.