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.

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.