Stoichiometry
Stoichiometric Calculations · Concentration and Dilution in Solution Stoichiometry
IB Chemistry topic guide
Reactivity 2: How Much, How Fast, How Far?
Reactivity 2: How Much, How Fast, How Far? is a core part of IB Chemistry. This guide connects the syllabus ideas behind Stoichiometry, Limiting Reagent and Yield, Rate of Reaction, Rate Laws and Mechanisms, How Far? The Extent of Chemical Change, shows how they appear in worked problems, and points you to the formulas and full lessons needed for exam revision.
What you will learn
Reactivity 2: How Much, How Fast, How Far? syllabus outline
The units below follow the structure used in the full Study to Learn course. Use the outline to identify exactly which idea needs attention, then work through the public example before continuing to the complete lesson path.
Limiting Reagent and Yield
Identifying the Limiting Reagent · Percentage Yield and Atom Economy
Rate of Reaction
Reaction Kinetics and Collision Theory · Measuring and Graphing Reaction Rates
Rate Laws and Mechanisms
Rate Laws: Relating Rate to Concentration · Reaction Mechanisms and the Rate-Determining Step
How Far? The Extent of Chemical Change
Chemical Equilibrium · Calculating Kᴄ from Equilibrium Data
Free worked preview
Stoichiometric Calculations
This complete preview comes from the Stoichiometry unit. It introduces the core language, shows the method in context, and gives you a real example of the lesson quality before you create an account.
Quantitative Chemistry from Balanced Equations
The balanced chemical equation is a recipe — it tells you exactly how many moles of each reactant combine and how many moles of each product form. Stoichiometry is the skill of following that recipe, converting between masses, volumes, and concentrations using the mole as the universal go-between. This lesson revisits and deepens the mole concept from Topic A4 with a focus on systematic calculation. After this lesson, you will be able to calculate masses, gas volumes, and solution volumes for any substance in a reaction given the quantity of any other substance and a balanced equation.
Stoichiometry is the calculation of quantities of reactants and products in chemical reactions, based on the mole ratios in a balanced chemical equation.
The General Method
- Write and balance the chemical equation.
- Convert the known quantity (mass, gas volume, solution volume × concentration) into moles using n = m/M, n = V/22.7 (gases at STP), or n = cV (solutions).
- Use the mole ratio from the balanced equation to find moles of the desired substance.
- Convert moles back into the quantity required (mass, volume, concentration).
The Central Role of the Mole Ratio
The coefficients in a balanced equation give the ratio in which substances react and are produced. In 2H₂ + O₂ → 2H₂O: 2 mol H₂ reacts with 1 mol O₂ to produce 2 mol H₂O. If you are given the amount of any one substance, you can find the amount of any other by multiplying by the appropriate ratio of coefficients.
Gases at STP
At standard temperature and pressure (0°C, 100 kPa), 1 mol of any gas occupies 22.7 dm³. This provides a direct link between gas volume and moles: n(gas) = V(dm³) / 22.7.
Worked Example: Mass to Gas Volume
2.50 g of calcium carbonate reacts with excess hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Calculate the volume of CO₂ gas produced at STP. (M(CaCO₃) = 100 g mol⁻¹)
Step 1 (mass → moles): n(CaCO₃) = 2.50 / 100 = 0.0250 mol.
Step 2 (mole ratio): CaCO₃ : CO₂ = 1 : 1 → n(CO₂) = 0.0250 mol.
Step 3 (moles → gas volume): V(CO₂) = 0.0250 × 22.7 = 0.568 dm³ (568 cm³) at STP.
Common ErrorAlways balance the equation first before using mole ratios. An unbalanced equation gives wrong ratios and wrong answers at every subsequent step of the calculation.
Reviewed by the Study to Learn editorial team · Updated 2026-08-04