CLASS 11–12 CHEMISTRY · SOME BASIC CONCEPTS OF CHEMISTRY

Stoichiometry and the limiting reagent

Divide the moles of each reactant by its coefficient in the balanced equation. The smallest value marks the limiting reagent, and it decides how much product forms.

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Worked examples

Example 1 · level 1

For the reaction CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, a mixture contains 0.25 mol of CaCO₃ and 1 mol of HCl. Which reactant is the limiting reagent?

HCl
Neither: they are in the exact mole ratio
Cannot be decided from moles alone
CaCO₃

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CaCO₃: 0.25 ÷ 1 = 0.25; HCl: 1 ÷ 2 = 0.5. The smaller value is for CaCO₃, so CaCO₃ is the limiting reagent.
HCl is in excess.

Answer: CaCO₃

Example 2 · level 2

Saanvi reacts 12 g of Mg with 64 g of O₂ according to 2Mg + O₂ → 2MgO. What mass of MgO is formed, in grams, correct to 2 decimal places? (Atomic masses: Mg = 24, O = 16)

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M(Mg) = 24 g/mol; M(O₂) = 2 × 16 = 32 g/mol.
n(Mg) = 12 ÷ 24 = 0.5 mol; n(O₂) = 64 ÷ 32 = 2 mol.
Mg: 0.5 ÷ 2 = 0.25; O₂: 2 ÷ 1 = 2. The smaller value is for Mg, so Mg is the limiting reagent.
Moles of MgO = 2 × 0.25 = 0.5 mol.
M(MgO) = 24 + 16 = 40 g/mol, so mass = 0.5 × 40 = 20 g.

Answer: 20 g

Example 3 · level 3

Vihaan reacts 12 g of Mg with 32 g of O₂ according to 2Mg + O₂ → 2MgO. What mass of the excess reactant is left over after the reaction is complete? Give the answer in grams, correct to 2 decimal places. (Atomic masses: Mg = 24, O = 16)

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M(Mg) = 24 g/mol; M(O₂) = 2 × 16 = 32 g/mol.
n(Mg) = 12 ÷ 24 = 0.5 mol; n(O₂) = 32 ÷ 32 = 1 mol.
Mg: 0.5 ÷ 2 = 0.25; O₂: 1 ÷ 1 = 1. The smaller value is for Mg, so Mg is the limiting reagent.
O₂ that reacts = 1 × 0.25 = 0.25 mol.
O₂ left = 1 − 0.25 = 0.75 mol = 0.75 × 32 = 24 g.

Answer: 24 g