IGCSE Chemistry Stoichiometry: A Worked Method
Stoichiometry connects the amounts of reactants and products in a balanced chemical equation. The key is to convert to moles before using the equation ratio. This example is written for practice and is not copied from an exam paper.
Balance first
Consider the decomposition reaction CaCO3 -> CaO + CO2. It is already balanced: one mole of calcium carbonate produces one mole of carbon dioxide. If the equation were not balanced, any ratio taken from it would be wrong.
Convert the given mass to moles
Suppose 10 g of CaCO3 reacts completely. Using relative atomic masses Ca = 40, C = 12 and O = 16, the relative formula mass of CaCO3 is 100 g/mol. Moles of CaCO3 = 10 / 100 = 0.10 mol. Write the unit next to this intermediate result.
Use the mole ratio
The balanced equation gives a 1:1 ratio between CaCO3 and CO2, so 0.10 mol of CaCO3 produces 0.10 mol of CO2. This ratio is about particles, not masses; do not assume the masses are equal.
Convert to the requested answer
The relative formula mass of CO2 is 44 g/mol. Mass of CO2 = 0.10 x 44 = 4.4 g. A quick check: the product mass is less than the original 10 g because the other product, CaO, contains some of that mass.
Try it yourself
If 25 g of CaCO3 decomposes completely, what mass of CO2 forms? Follow the same four steps before checking your answer: 11 g. If your answer differs, first check the formula masses, then the mole ratio, then the final conversion.
What to do next
When questions include a limiting reactant, calculate the amount of product possible from each reactant and choose the smaller amount. Practise this only after the one-reactant method feels automatic.