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. M_r of C = 12g M_r of O₂ = 32g	1. Find the Relative Formula Mass of the reactants. M_r of NaOH = 40g M_r of HCl = 36.5g	1. Find the Relative Formula Mass of the reactants. M_r of NaOH = 40g M_r 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.

Limiting Reactants and Concentration
If 0.1 dm³ of 0.5M Hydrochloric Acid is added to 0.1 dm³ of 0.6M Sodium Hydroxide which is the limiting reactant in this reaction?	If 0.12 dm³ of 0.5M Sulphuric Acid is added to 0.1 dm³ of 1.1M Sodium Hydroxide which is the limiting reactant in this reaction?	If 0.1 dm³ of 1.5M Hydrochloric Acid is added to 0.300 dm³ of 0.2M Magnesium Hydroxide which is the limiting reactant in this reaction?
1. Write the Balanced Symbol Equation for the reaction. <chem>HCl+NaOH->NaCl+H2O</chem>	1. Write the Balanced Symbol Equation for the reaction. <chem>H2SO4+2NaOH->Na2SO4+2H2O</chem>	1. Write the Balanced Symbol Equation for the reaction. <chem>2HCl+Mg(OH)2->MgCl2+2H2O</chem>
2. State the ratio of moles of each chemical needed. 1 moles of HCl is needed for every 1 mole of NaOH	2. State the ratio of moles of each chemical needed. 1 moles of H₂SO₄ are needed for every 2 moles of NaOH	2. State the ratio of moles of each chemical needed. 2 moles of HCl is needed for every 1 mole of Mg(OH)₂
3. Find the number of moles supplied of each chemical. concentration of HCl = \(\frac{Moles (mol)}{volume (dm^3)}\) 0.5 = \(\frac{Moles}{0.1}\) Moles of HCl = 0.05 mol Therefore 0.05 mol of NaOH are needed. concentration of NaOH = \(\frac{Moles (mol)}{volume (dm^3)}\) 0.6 = \(\frac{Moles}{0.1}\) Moles of NaOH = 0.06 mol There is more than enough NaOH, therefore HCl is the limiting reactant.	3. Find the number of moles supplied of each chemical. concentration of H₂SO₄ = \(\frac{Moles (mol)}{volume (dm^3)}\) 0.5 = \(\frac{Moles}{0.12}\) Moles of H₂SO₄ = 0.06 mol Therefore 0.12 mol of NaOH are needed. concentration of NaOH = \(\frac{Moles (mol)}{volume (dm^3)}\) 1.1 = \(\frac{Moles}{0.1}\) Moles of NaOH = 0.11 mol There is not enough NaOH, therefore NaOH is the limiting reactant.	3. Find the number of moles supplied of each chemical. concentration of HCl = \(\frac{Moles (mol)}{volume (dm^3)}\) 1.5 = \(\frac{Moles}{0.1}\) Moles of HCl = 0.15 mol Therefore 0.075 mol of Mg(OH)₂ are needed. concentration of Mg(OH)₂ = \(\frac{Moles (mol)}{volume (dm^3)}\) 0.2 = \(\frac{Moles}{0.3}\) Moles of Mg(OH)₂ = 0.06 mol There is not enough Mg(OH)₂, therefore Mg(OH)₂ is the limiting reactant.