Limiting & Excess Reactants

Excess & limiting reactants

Sometimes, there is an excess of one or more of the reactants (excess reactant)
The reactant which is not in excess is called the limiting reactant
To determine which reactant is limiting:
- The number of moles of the reactants should be calculated
- The ratio of the reactants shown in the equation should be taken into account eg:

C + 2H₂ → CH₄

What is limiting when 10 mol of carbon are reacted with 3 mol of hydrogen?
- Hydrogen is the limiting reactant and since the ratio of C : H₂ is 1:2 only 1.5 mol of C will react with 3 mol of H₂

Worked example

9.2 g of sodium metal is reacted with 8.0 g of sulfur to produce sodium sulfide, Na₂S.Which reactant is in excess and which is limiting?

Answer:

Step 1: Calculate the moles of each reactant

$number of moles (Na) = \frac{9.2 g}{22.99 g {mol}^{- 1}} = 0.40 mol$

$number of moles (S) = \frac{8.0 g}{32.07 g {mol}^{- 1}} = 0.25 mol$

Step 2: Write the balanced equation and determine the coefficients

2Na + S → Na₂S

Step 3: Divide the moles by the coefficient and determine the limiting reagent

- divide 0.40 moles of Na by 2, giving 0.20 - lowest
- divide 0.25 moles of S by 1, giving 0.25

Therefore, sodium is limiting and sulfur is in excess

Exam Tip

An easy way to determine the limiting reactant is to find the moles of each substance and divide the moles by the coefficient in the equation. The lowest number resulting is the limiting reactant

In the example above:
- divide 10 moles of C by 1, giving 10
- divide 3 moles of H by 2, giving 1.5, so hydrogen is limiting

Limiting & Excess Reactants (HL IB Chemistry)

Revision Note