This mini-lesson covers Reactivity 2.1: mole ratios from balanced equations, limiting reagents, percentage yield and atom economy. This HL lesson also builds in the Additional Higher Level (AHL) material.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Press Start when you're ready.
A balanced equation conserves mass and gives the mole ratio in which substances react. For N₂ + 3H₂ → 2NH₃, one mole of N₂ reacts with three of H₂ to give two of NH₃.
Strategy: convert what you know to moles, apply the ratio, then convert back to mass, volume or concentration.
To find a product mass: moles of known → mole ratio → moles of product → mass (m = n × M).
Example: how much ammonia from 1 mol N₂ (with plenty of H₂)? Ratio 1 : 2 → 2 mol NH₃ → mass = 2 × 17.03 g mol⁻¹.
The limiting reagent is the reactant that runs out first — it caps how much product can form. The other reactant is in excess. Compare moles ÷ coefficients to find which is limiting.
For Mg + 2HCl → MgCl₂ + H₂, each mole of Mg needs two moles of HCl.
The theoretical yield is the maximum from the equation. Reactions rarely reach it (side reactions, losses, reversible reactions), so:
Atom economy measures how much of the reactant mass ends up in the useful product — a key idea in green chemistry:
In a titration the moles of one solution are found from n = C × V, then the balanced equation gives the moles (and concentration) of the other. Always convert cm³ to dm³ first.
Tap a phrase, then the idea it describes.
Tap an item on the left, then its match on the right.
Mole ratios: balance the equation, convert to moles, apply the ratio
Limiting reagent: runs out first and caps the product; the other is in excess
Yield: % yield = (actual ÷ theoretical) × 100
Atom economy: (useful product M_r ÷ total M_r) × 100 — greener when higher
You've covered How much? — stoichiometry for IB Diploma Chemistry HL. Press Finish to see your score.
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