This mini-lesson covers Reactivity 2.1: mole ratios from balanced equations, limiting reagents, percentage yield and atom economy.
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:
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 SL. Press Finish to see your score.
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