Limiting Reactant

Key Stage 4

Meaning

A limiting reactant is a reactant that is not supplied in a large enough quantity for a complete reaction with the other reactants.

About Limiting Reactants

A limiting reactant prevents a complete reaction.

Examples

2H₂(g) + O₂(g) → 2H₂O(l)

If there is 4g of Hydrogen then 32g of Oxygen is needed for a complete reaction to occur.

If there is 4g of Hydrogen but only 31g of Oxygen then the Oxygen is a limiting reactant.

If there is 3g of Hydrogen but only 32g of Oxygen then the Hydrogen is a limiting reactant.

Finding a Limiting Reactant

Limiting Reactants and Mass

If there are are 6g of Carbon and 15g of Oxygen in the following reaction: C + O2 → CO2 Which is the limiting reactant in this reaction?	If there are are 27g of Aluminium and 32g of Oxygen in the following reaction: 4Al + 3O2 → 2Al2O3 Which is the limiting reactant in this reaction?	If there are are 180g of Glucose and 200g of Oxygen in the following reaction: C6H12O6 + 6O2 → 6H2O + 6CO2 Which is the limiting reactant in this reaction?
1. Find the Relative Formula Mass of the reactants. Mr of C = 12g Mr of O2 = 32g	1. Find the Relative Formula Mass of the reactants. Mr of NaOH = 40g Mr of HCl = 36.5g	1. Find the Relative Formula Mass of the reactants. Mr of NaOH = 40g Mr of HCl = 36.5g
2. State the ratio of moles of each chemical needed. 1 moles of C is needed for every 1 mole of O2	2. State the ratio of moles of each chemical needed. 4 moles of Al are needed for every 3 moles of O2	2. State the ratio of moles of each chemical needed. 1 mole of C6H12O6 is needed for every 6 moles of O2
3. Find the number of moles supplied of each chemical. No. Moles of C = \(\frac{Mass}{M_r}\) No. Moles of C = \(\frac{6}{12}\) No. Moles = 0.5 mol Therefore 0.5 moles of O2 is needed. No. Moles of O = \(\frac{Mass}{M_r}\) No. Moles of O = \(\frac{15}{32}\) No. Moles = 0.46 mol Oxygen is the limiting reactant. 1 mole of O2 = 32g	3. Find the number of moles supplied of each chemical. No. Moles of Al = \(\frac{Mass}{M_r}\) No. Moles of Al = \(\frac{27}{27}\) No. Moles = 1 mol Therefore 0.75 moles of O2 are needed. No. Moles of O = \(\frac{Mass}{M_r}\) No. Moles of O = \(\frac{32}{32}\) No. Moles = 1 mol Oxygen is the limiting reactant.	3. Find the number of moles supplied of each chemical. No. Moles of C6H12O6 = \(\frac{Mass}{M_r}\) No. Moles of C6H12O6 = \(\frac{180}{180}\) No. Moles = 1 mol Therefore 6 moles of O2 are needed. No. Moles of O = \(\frac{Mass}{M_r}\) No. Moles of O = \(\frac{200}{32}\) No. Moles = 6.25 mol Glucose is the limiting reactant.