Preview

Stoichiometry

Q: 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.

Q: 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.

Q: 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%.

Mole ratios, limiting reagents, and theoretical yield

A recipe with a fatal flaw: it never tells you when to stop. Three moles of hydrogen, two of oxygen — looks like plenty of everything. But the reaction has its own arithmetic: it eats hydrogen twice as fast as oxygen, and it stops dead the instant one ingredient runs out, no matter how much of the other is still standing there. Chemistry doesn't care how much you brought — only how it's portioned. Today you learn to spot the ingredient that calls the shots.

What you'll be able to do

Use mole ratios from a balanced equation to convert between amounts of reactants and products
Identify the limiting reagent by comparing moles-per-coefficient, and compute theoretical yield from it
Verify conservation of mass/atoms before and after a reaction at the molecular level

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

Make a prediction

For 2H₂ + O₂ → 2H₂O, you load 3 mol of H₂ and 2 mol of O₂. Which reactant runs out first?

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Your prediction

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Answer: H₂ — each reaction event eats 2 H₂ per 1 O₂, so 3 mol of H₂ only covers 1.5 mol of O₂

The reaction counts portions, not headcounts. Per the equation, every O₂ needs 2 H₂. Compare moles-per-coefficient: H₂ has 3/2 = 1.5, O₂ has 2/1 = 2 — hydrogen's portion runs out first, leaving 0.5 mol of O₂ standing unused. The product: 1.5 × 2 = 3 mol of H₂O. 'Fewer moles' is not 'runs out first', and reactant moles never simply add into product moles — the coefficients do the bookkeeping.

Quiz (0/3)

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

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

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

You can now

Use mole ratios from a balanced equation to convert between amounts of reactants and products
Identify the limiting reagent by comparing moles-per-coefficient, and compute theoretical yield from it
Verify conservation of mass/atoms before and after a reaction at the molecular level