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 at a ratio of 1:1 then 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 at a ratio of 4:3 then 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 at a ratio of 1:6 then 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 at a ratio of 1:1 then 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 at a ratio of 1:2 then 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 at a ratio of 2:1 then 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.

Limiting Reactants and Volume

If 12 dm3 of Hydrogen gas reacts with 15 dm3 Fluorine gas which is the limiting reactant in this reaction?	If 18 dm3 of Hydrogen gas reacts with 8 dm3 Oxygen gas which is the limiting reactant in this reaction?	If 9 dm3 of Ethane reacts with 32 dm3 of Oxygen which is the limiting reactant in this reaction?
1. Write the Balanced Symbol Equation for the reaction. <chem>H2 + F2->2HF</chem>	1. Write the Balanced Symbol Equation for the reaction. <chem>2H2 + O2-> 2H2O</chem>	1. Write the Balanced Symbol Equation for the reaction. <chem>2C2H6 + 7O2->4CO2 + 6H2O</chem>
2. State the ratio of moles of each chemical needed. 1 moles of H2 is needed for every 1 mole of F2	2. State the ratio of moles of each chemical needed. 2 moles of H2 is needed for every 1 mole of O2	2. State the ratio of moles of each chemical needed. 2 moles of C2H6 is needed for every 7 moles of O2
3. Find the number of moles supplied of each chemical. Volume of H2 = 24 x (number of moles) 12 = 24 x (number of moles) Moles of H2 = 12/24 Moles of H2 = 0.5 mol Therefore at a ratio of 1:1 then 0.5 mol of F2 are needed. Volume of F2 = 24 x (number of moles) 15 = 24 x (number of moles) Moles of F2 = 15/24 Moles of F2 = 0.625 mol There is more than enough F2, therefore H2 is the limiting reactant.	3. Find the number of moles supplied of each chemical. Volume of H2 = 24 x (number of moles) 18 = 24 x (number of moles) Moles of H2 = 18/24 Moles of H2 = 0.75 mol Therefore at a ratio of 2:1 then 0.375 mol of O2 are needed. Volume of O2 = 24 x (number of moles) 8 = 24 x (number of moles) Moles of O2 = 8/24 Moles of O2 = 0.333 mol There is not enough O2, therefore O2 is the limiting reactant.	3. Find the number of moles supplied of each chemical. Volume of H2 = 24 x (number of moles) 9 = 24 x (number of moles) Moles of C2H6 = 9/24 Moles of C2H6 = 0.375 mol Therefore at a ratio of 2:7 then 1.3125 mol of O2 are needed. Volume of O2 = 24 x (number of moles) 32 = 24 x (number of moles) Moles of O2 = 32/24 Moles of O2 = 1.333 mol There is more than enough O2, therefore C2H6 is the limiting reactant.

References

AQA

Limiting reactants, page 45, GCSE Chemistry; The Revision Guide, CGP, AQA

Limiting reactants, page 67, GCSE Chemistry; Third Edition, Oxford University Press, AQA

Limiting reactants, pages 114-116, GCSE Combined Science Trilogy; Chemistry, CGP, AQA

Limiting reactants, pages 124-126, GCSE Chemistry, CGP, AQA

Limiting reactants, pages 190-1, GCSE Combined Science Trilogy 1, Hodder, AQA

Limiting reactants, pages 75-6, GCSE Chemistry, Hodder, AQA

Edexcel

Limiting reactant, page 220, GCSE Combined Science, Pearson Edexcel

Limiting reactant, page 76, GCSE Chemistry, Pearson, Edexcel

Limiting reactants, page 31, GCSE Chemistry; The Revision Guide, CGP, Edexcel

Limiting reactants, page 84, GCSE Chemistry, CGP, Edexcel

Limiting reactants, page 94, GCSE Combined Science; The Revision Guide, CGP, Edexcel

OCR

Limiting reactants, pages 39, 40, Gateway GCSE Chemistry; The Revision Guide, CGP, OCR

Limiting reactants, pages 98, 158, Gateway GCSE Chemistry, Oxford, OCR