Stoichiometry & Limiting Reactants
- Use mole ratios from a balanced equation to convert between reactants and products
- Identify the limiting reactant and calculate the theoretical yield
- Compute percent yield from actual and theoretical yields
The mole ratio is the heart of stoichiometry
A balanced equation is a recipe in moles. In N₂ + 3H₂ → 2NH₃, the coefficients say 1 mole of N₂ reacts with 3 moles of H₂ to make 2 moles of NH₃. These mole ratios (like 3 mol H₂ : 1 mol N₂) are the conversion factors that carry you from one substance to another. Grams never react in whole-number ratios — moles do — so you always route through moles.
Limiting reactant and percent yield
When reactants are not supplied in the exact recipe ratio, one runs out first — the limiting reactant — and it caps how much product can form. The other is in excess. To find it, convert each reactant to moles and compare against the balanced ratio; whichever produces the least product is limiting. The amount of product it predicts is the theoretical yield. Reactions are rarely perfect, so percent yield measures how close reality came.
For N₂ + 3H₂ → 2NH₃, you start with 28 g of N₂ (molar mass 28 g·mol⁻¹) and 10 g of H₂ (molar mass 2 g·mol⁻¹). Find the limiting reactant and the mass of NH₃ (17 g·mol⁻¹) formed.
- 1.Convert to moles: N₂ = 28 g ÷ 28 g·mol⁻¹ = 1.0 mol; H₂ = 10 g ÷ 2 g·mol⁻¹ = 5.0 mol.
- 2.Use the 1 N₂ : 3 H₂ ratio. To react all 1.0 mol N₂ you need 3.0 mol H₂ — you have 5.0 mol, so H₂ is in excess and N₂ is limiting.
- 3.Product from the limiting reactant: 1.0 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 2.0 mol NH₃.
- 4.Convert to mass: 2.0 mol × 17 g·mol⁻¹ = 34 g NH₃. (Leftover H₂: 5.0 − 3.0 = 2.0 mol = 4 g.)
The reaction above has a theoretical yield of 34 g NH₃, but you actually collect 25.5 g. What is the percent yield?
- 1.percent yield = (actual ÷ theoretical) × 100%.
- 2.= (25.5 g ÷ 34 g) × 100%.
- 3.= 0.75 × 100%.
Never pick the limiting reactant by mass or by "whichever there is less of." 10 g of H₂ is far fewer grams than 28 g of N₂, yet H₂ is the one in excess. Always compare in moles against the balanced ratio.
For 2H₂ + O₂ → 2H₂O, you have 4 mol H₂ and 3 mol O₂. Which is the limiting reactant?
For 2Al + 3Cl₂ → 2AlCl₃, how many moles of Cl₂ are required to react completely with 4 mol of Al?
On limiting-reactant free-response, show the moles-of-product each reactant would make and explicitly state which is smaller. The reactant giving the smaller product amount is limiting, and that smaller number is your theoretical yield.
Answer the 2 checkpoints as you read.
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