Use a balanced equation to calculate a mass of reactant or product.
The chemistry
Combine each balanced coefficient with the relative mass of its substance to find the corresponding reacting masses. Scale these masses in the same ratio.
At Higher tier, you can also divide the given mass by its molar mass, apply the mole ratio, then multiply by the product molar mass. Edexcel Foundation uses the mass-ratio route here; the mole route is Higher-tier working.
Both routes use the same chemistry. The task assumes the stated reactant reacts completely and any other reactant is in excess.
Worked example
2 Mg + O₂ → 2 MgO. With Mg = 24 and O = 16, MgO has Mr = 40.
The mass-ratio route gives 4.8 × 40 ÷ 24 = 8.00 g. The extra mass comes from oxygen.
Higher-tier mole route: 4.8 g Mg gives 4.8 ÷ 24 = 0.20 mol Mg, then 0.20 mol MgO, giving 8.00 g MgO.
The stated reaction goes to completion. Any other reactant is in excess; all of the stated reactant reacts. Relative atomic masses: H = 1, C = 12, O = 16, Mg = 24, Ca = 40.
1 CaCO3 → 1 CaO + 1 CO2
What mass of calcium oxide forms from 15 g of calcium carbonate? Give your answer to 2 decimal places.
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