For 2H₂ + O₂ → 2H₂O, with 3 mol H₂ and 2 mol O₂, which is limiting?

H₂: 3/2 = 1.5, O₂: 2/1 = 2. H₂ has the smaller ratio → H₂ is limiting.

For N₂ + 3H₂ → 2NH₃, starting with 1 mol N₂ and 4 mol H₂, how many moles of NH₃ form?

N₂: 1/1=1, H₂: 4/3=1.33 → N₂ is limiting. NH₃ = 1 × (2/1) = 2 mol.

If theoretical yield is 5.0 g but you obtain 3.8 g, what is the percent yield?

% yield = (3.8/5.0) × 100% = 76%.

Pro 🔒~20 min

Stoichiometry

Mole ratios, limiting reagents, and theoretical yield

How it works

Stoichiometry uses the coefficients of a balanced chemical equation to relate amounts of reactants and products. The mole ratio (coefficient ratio) converts between moles of different substances. The limiting reagent is the reactant that runs out first, determining the maximum product (theoretical yield). The excess reagent has leftover moles after the reaction completes. For a reaction aA + bB → cC: if (moles A)/a < (moles B)/b, then A is limiting. Theoretical moles of C = (moles of limiting reagent) × (c/a or c/b). Conservation of mass requires that total atoms of each element are equal on both sides.

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Step-by-step

Choose a reaction type and adjust the moles of each reactant.
The simulation displays molecules as colored circles — watch them combine and identify which reactant runs out first (the limiting reagent).
The data panel shows mole ratios, theoretical yield, and leftover excess in real time.

Key formulas

$\text{moles of product} = \text{moles of limiting reagent} \times \frac{\text{product coeff.}}{\text{limiting reagent coeff.}}$ Theoretical yield is determined by the limiting reagent and the mole ratio from the balanced equation
$\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$ Percent yield compares actual product obtained to the theoretical maximum

Frequently asked questions

For 2H₂ + O₂ → 2H₂O, with 3 mol H₂ and 2 mol O₂, which is limiting?: H₂: 3/2 = 1.5, O₂: 2/1 = 2. H₂ has the smaller ratio → H₂ is limiting.
For N₂ + 3H₂ → 2NH₃, starting with 1 mol N₂ and 4 mol H₂, how many moles of NH₃ form?: N₂: 1/1=1, H₂: 4/3=1.33 → N₂ is limiting. NH₃ = 1 × (2/1) = 2 mol.
If theoretical yield is 5.0 g but you obtain 3.8 g, what is the percent yield?: % yield = (3.8/5.0) × 100% = 76%.