← Back to subjects
0
IB Diploma Chemistry SL · Reactivity 2.1 — the amount of chemical change
Mini-Lesson

How much? — stoichiometry

This mini-lesson covers Reactivity 2.1: mole ratios from balanced equations, limiting reagents, percentage yield and atom economy.

mole ratioslimiting reagentyield & economy Reactivity 2.1 — the amount of chemical change

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.

Reactivity 2.1

Balanced equations and mole ratios

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.

Reactivity 2.1

Mass calculations

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.

Calculate

Mass of product

1What mass of NH₃ (M = 17.03) forms from 1.00 mol N₂ reacting completely (N₂ + 3H₂ → 2NH₃)?
g
Ratio 1 : 2 gives 2.00 mol NH₃; mass = 2.00 × 17.03.
Reactivity 2.1

Limiting reagent

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.

Calculate

Using the limiting reagent

20.10 mol Mg reacts with 0.15 mol HCl (Mg + 2HCl → MgCl₂ + H₂). HCl is limiting. How many moles of H₂ form?
mol
HCl : H₂ is 2 : 1, so n(H₂) = 0.15 ÷ 2.
Reactivity 2.1

Percentage yield

The theoretical yield is the maximum from the equation. Reactions rarely reach it (side reactions, losses, reversible reactions), so:

% yield = (actual ÷ theoretical) × 100compares what you got to the maximum possible
Calculate

Percentage yield

3A reaction should give 8.0 g of product but only 6.4 g is collected. Calculate the percentage yield.
%
% yield = (6.4 ÷ 8.0) × 100.
Reactivity 2.1

Atom economy

Atom economy measures how much of the reactant mass ends up in the useful product — a key idea in green chemistry:

atom economy = (Mr of useful product ÷ Mr of all products) × 100higher = less waste
Calculate

Atom economy

4For CaCO₃ → CaO + CO₂, the useful product is CaO (56.08). Total product mass = CaO + CO₂ = 100.09. Calculate the atom economy for CaO.
%
(56.08 ÷ 100.09) × 100.
Quick check

Why less than 100%?

?A percentage yield below 100% is commonly due to:
Sort it

Reagent roles and yield

Tap a phrase, then the idea it describes.

🎯 Limiting reagent

➕ Excess reagent

📉 Lowers % yield

Match it

Match term to formula

Tap an item on the left, then its match on the right.

Term
Definition / formula
Recap

The big ideas to know

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.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through How much? — stoichiometry for IB Diploma Chemistry SL. 🎉

Your stars: 0 / 0

Next: test yourself in the Evaluate stage Confidence Quiz, then lock it in with Verify.

📣 Smashed it? Share your score

Challenge a mate to beat your stars, or show a parent how you got on.

→ Back to all subjects