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 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