The Limiting Reactant

Part of the Stoichiometry section of Coddy's Chemistry journey. Lesson 25 of 41.

A balanced equation gives the mole ratio in which the reactants are used. In any other ratio one of them runs out first: the limiting reactant. When it is gone the reaction stops, so it decides how much product forms.

To find it, turn each amount into moles and divide by its coefficient: the smallest result wins. In 2H₂ + O₂ → 2H₂O with 3 mol H₂ and 1 mol O₂: H₂ gives 1.5, O₂ gives 1. Oxygen is limiting.

Two containers of gas: one with 18 hydrogen particles labeled 3 mol, one with 6 oxygen particles labeled 1 mol

3 mol H₂ and 1 mol O₂: which runs out first?

Work out the product from the limiting reactant only: 1 mol O₂ makes 2 mol H₂O. The hydrogen that is not needed is left over.

Challenge

A strip of magnesium, $4.8\ \mathrm{g}$ , burns in a sealed jar that holds $4.0\ \mathrm{g}$ of oxygen: 2Mg + O₂ → 2MgO. What mass of magnesium oxide, MgO, forms?