Limiting Reactant

Limiting reactant is the reactant that completely reacts and determine the amount of products form.
The substance that remains after a reaction is completed is the excess reactant.

Example 1:

2.3g of sodium is heated with 6.4g of oxygen to form sodium oxide.
Explain which substance is the limiting reactant and determine the mass of sodium oxide formed.

Mole of sodium = mass /Ar = 2.3/23 = 0.1mol
Mole of oxygen = 6.4/32 = 0.2mol

Balanced equation: 2Na + O₂ → 2Na₂O

According to balanced chemical equation, 2 mol of sodium reacts with 1mol of oxygen. 0.1mol of sodium reacts with 0.05mol of oxygen.
We are provided with 0.2mol of Oxygen which is more than 0.05mol so oxygen is in excess, and Sodium is the limiting reactant.

Using mole of limiting reactant to determine mole of the product

2 mol of sodium produces 2 mol of sodium oxide.
0.1 mol of sodium produces 0.1mol of sodium oxide.
Mass of sodium oxide = mole X Mr = 0.1 x 62 = 6.2g

Example 2:

10.8g of Al is mixed with 7.2dm³ of fluorine in a container at room temperature and pressure.
The container is heated, and aluminium fluoride is formed. Find the mass of aluminium fluoride formed.

Mole of Al = 10.8/27 = 0.4mol
Mole of fluorine= 7.2/24 = 0.3mol

Balanced equation: 2Al + 3F₂ → 2AlF₃

According to balanced chemical equation, 2 mol of aluminium reacts with 3 mol of fluorine. 0.2mol of aluminium reacts with 0.3mol of fluorine.
We are provided with 0.4mol of aluminium which is more than 0.2mol thus aluminium is in excess, and fluorine is the limiting reactant.

Using mole of limiting reactant to determine mole of the product

3 mol of fluorine produces 2 mol of aluminium fluoride.
0.3mol of fluorine will produce 0.2 mol of aluminium fluoride.
Mass of aluminium fluoride = mole x Mr = 0.2 x 84 = 16.8g