Limiting Reagents and Reactants in Excess Chemistry Tutorial

Key Concepts

• Some chemical reactions go to completion.⁽¹⁾

  chemical reaction
  reactants			→	products
  aA	+	bB	→	cC	+	dD

• The limiting reagent is the reactant that is completely used up during the chemical reaction.

  chemical reaction
  reactants			→	products
  aA	+	bB	→	cC	+	dD

  If reactant A is the limiting reagent, moles of A left over on completion will be 0 (n(A) = 0 mol)

  If reactant B is the limiting reagent, moles of B left over on completion will be 0 (n(B) = 0 mol)

• The reactant in excess is the reactant that is not completely used up during the chemical reaction, that is, there is some of this reactant left over.

  chemical reaction
  reactants			→	products
  aA	+	bB	→	cC	+	dD

  If reactant A is the reactant in excess, some moles of A will be left over on completion (n(A) > 0 mol)

  If reactant B is the reactant in excess, some moles of B will be left over on completion (n(B) > 0 mol)

• Deciding which reactants are the limiting reagents and the reactants in excess:

1. Write the balanced chemical equation for the chemical reaction
2. Calculate the available moles (n) of each reactant in the chemical reaction

   For masses: n = m ÷ M
   For solutions:⁽²⁾ n = c × V
   For gases at STP: n = V ÷ 22.71
   For gases at SLC (SATP):⁽³⁾ n = V ÷ 24.79

3. Use the balanced chemical equation to determine the mole ratio (stoichiometric ratio) of the reactants in the chemical reaction
4. Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio
5. (i) The limiting reagent is the reactant that will be completely used up during the chemical reaction.

   n(limiting reagent) = 0
   on completion of reaction

   (ii) There will be some moles of the reactant in excess left over after the reaction has gone to completion.

   n(reactant in excess) > 0
   on completion of reaction

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Limiting Reagents and Reactants in Excess Example: moles of reactants given

Question: Find the limiting reagent and the reactant in excess when 0.5 moles of Zn react completely with 0.4 moles of HCl

Solution:

1. Write the balanced chemical equation for the chemical reaction

   Zn + 2HCl → ZnCl₂ + H₂

2. Calculate the available moles of each reactant in the chemical reaction

   Calculate moles of Zn available to react moles of Zn = 0.5 mol

   Calculate moles of HCl available to react moles of HCl = 0.4 mol

3. Use the balanced chemical equation to determine the mole ratio (stoichiometric ratio) of the reactants in the chemical reaction

   Zn : HCl	or	HCl : Zn
   1 : 2		1 : ½

4. Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio

   If all of the 0.5 moles of Zn were to be used in the reaction it would require

   2 × 0.5 = 1.0 moles of HCl for the reaction to go to completion.

   There are only 0.4 moles of HCl available which is less than the required 1.0 moles.

   If all of the 0.4 moles of HCl were to be used in the reaction it would require

   ½ × 0.4 = 0.2 moles Zn.

   There are 0.5 moles of Zn available which is more than the required 0.2 moles.

5. (i) The limiting reagent is the reactant that will be completely used up during the chemical reaction.
   (ii) There will be some moles of the reactant in excess left over after the reaction has gone to completion.

   The limiting reagent is HCl
   (all of the 0.4 moles of HCl will be used up when this reaction goes to completion)

   The reactant in excess is Zn
   (when the reaction has gone to completion there will be 0.5 - 0.2 = 0.3 moles of Zn left over)

Limiting Reagents and Reactants in Excess Example: masses of reactants given

Question: Find the limiting reagent and the reactant in excess when 1.5 g of CaCO₃ react completely with 0.73 g of HCl

Solution:

1. Write the balanced chemical equation for the chemical reaction

   CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

2. Calculate the available moles of each reactant in the chemical reaction

   Calculate moles of CaCO3 available to react moles of CaCO3 = mass ÷ molar mass mass = 1.5 g molar mass = 40.08 + 12.01 + (3 × 16.00) = 100.09 g mol-1 moles of CaCO3 = 1.5 ÷ 100.09 = 0.015 mol

   Calculate moles of HCl available to react moles of HCl = mass ÷ molar mass mass = 0.73 g molar mass = 1.008 + 35.45 = 36.458 g mol-1 moles HCl = 0.73 ÷ 36.458 = 0.02 mol

3. Use the balanced chemical equation to determine the mole ratio (stoichiometric ratio) of the reactants in the chemical reaction

   CaCO3 : HCl	or	HCl : CaCO3
   1 : 2		1 : ½

4. Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio

   If all of the 0.015 moles of CaCO₃ were to be used in the reaction it would require

   2 × 0.015 = 0.03 moles of HCl for the reaction to go to completion.

   There are only 0.02 moles of HCl available which is less than the required 0.03 moles.

   If all of the 0.02 moles of HCl were to be used in the reaction it would require

   ½ × 0.02 = 0.01 moles of CaCO₃.

   There are 0.015 moles of CaCO₃ available which is more than the required 0.01 moles.

5. (i) The limiting reagent is the reactant that will be completely used up during the chemical reaction.
   (ii) There will be some moles of the reactant in excess left over after the reaction has gone to completion.

   The limiting reagent is HCl
   (all of the 0.02 moles of HCl will be used up when this reaction goes to completion)

   The reactant in excess is CaCO₃
   (when the reaction has gone to completion there will be 0.015 - 0.01 = 0.005 moles of CaCO₃ left over)

Limiting Reagents and Reactants in Excess Example: concentration and volume of solutions given

Question: Find the limiting reagent and the reactant in excess when 100 mL of 0.2 mol L^-1 NaOH aqueous solution react completely with 50 mL of 0.5 mol L^-1 H₂SO₄ aqueous solution.

Solution:

1. Write the balanced chemical equation for the chemical reaction

   2NaOH_(aq) + H₂SO_4(aq) → Na₂SO_4(aq) + 2H₂O_(l)

2. Calculate the available moles of each reactant in the chemical reaction:

   n = c × V

   n = moles of solute in mol
   c = concentration of solution (molarity) in mol L^-1
   V = volume of solution in L
   Note: 1 L = 10³ mL (or 1000 mL) so 1 mL = 10^-3 L (or 0.001 L)

   Calculate moles of NaOH available to react n(NaOH) = c(NaOH) × V c(NaOH) = 0.2 mol L-1 V = 100 mL = 100 × 10-3 L (or 0.100 L) n(NaOH) = 0.2 × (100 × 10-3) = 0.02 mol or n(NaOH) = 0.2 × 0.100 = 0.02 mol

   Calculate moles of H2SO4 available to react n(H2SO4) = c(H2SO4) × V c(H2SO4) = 0.5 mol L-1 V = 50 mL = 50 × 10-3 L (or 0.050 L) n(H2SO4) = 0.5 × (50 × 10-3) = 0.025 mol or n(H2SO4) = 0.5 × 0.050 = 0.025 mol

3. Use the balanced chemical equation to determine the mole ratio (stoichiometric ratio) of the reactants in the chemical reaction

   NaOH : H2SO4	or	H2SO4 : NaOH
   1 : ½		1 : 2

4. Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio

   If all of the 0.02 moles of NaOH were to be used in the reaction it would require

   ½ × 0.02 = 0.01 moles of H₂SO₄ for the reaction to go to completion.

   There are 0.025 moles of H₂SO₄ available which is more than the required 0.01 moles.

   If all of the 0.025 moles of H₂SO₄ were to be used in the reaction it would require

   2 × 0.025 = 0.05 moles of NaOH.

   There are only 0.02 moles of NaOH available which is less than the required 0.05 moles.

5. (i) The limiting reagent is the reactant that will be completely used up during the chemical reaction.
   (ii) There will be some moles of the reactant in excess left over after the reaction has gone to completion.

   The limiting reagent is NaOH
   (all of the 0.02 moles of NaOH will be used up when this reaction goes to completion)

   The reactant in excess is H₂SO₄
   (when the reaction has gone to completion there will be 0.025 - 0.01 = 0.015 moles of H₂SO₄ left over)

Limiting Reagents and Reactants in Excess Example: gas volumes given

Question: Find the limiting reagent and the reactant in excess when 45.42 L of CO(g) react completely with 11.36 L of O₂(g) at STP (0°C or 273.15 K and 100 kPa)

Solution:

1. Write the balanced chemical equation for the chemical reaction

   2CO(g) + O₂(g) → 2CO₂(g)

2. Calculate the available moles of each reactant in the chemical reaction

   Calculate moles of CO(g) available to react n(CO(g)) = V(CO(g)) ÷ Vm At STP. 1 mole of gas has a volume of 22.71 L Vm = 22.71 L mol-1 V(CO(g)) = 45.42 L moles of CO = 45.42 ÷ 22.71 = 2 mol

   Calculate moles of O2(g) available to react n(O2) = V(O2(g) ÷ Vm At STP 1 mole of gas has a volume of 22.71 L Vm = 22.71 L mol-1 V(O2(g)) = 11.36 L moles O2 = 11.36 ÷ 22.71 = 0.5 mol

3. Use the balanced chemical equation to determine the mole ratio (stoichiometric ratio) of the reactants in the chemical reaction

   CO : O2	or	O2 : CO
   1 : ½		1 : 2

4. Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio

   If all of the 2 moles of CO were to be used in the reaction it would require

   ½ × 2 = 1 mole of O₂ for the reaction to go to completion.

   There are 0.5 moles of O₂ available which is less than the required 1 mole.

   If all of the 0.5 moles of O₂ were to be used in the reaction it would require

   2 × 0.5 = 1 mole of CO.

   There are 2 moles of CO available which is more than the required 1 mole.

5. (i) The limiting reagent is the reactant that will be completely used up during the chemical reaction.

   (ii) There will be some moles of the reactant in excess left over after the reaction has gone to completion.

   The limiting reagent is O₂
   (all of the 0.5 moles of O₂ will be used up when this reaction goes to completion)

   The reactant in excess is CO
   (when the reaction has gone to completion there will be 2 - 1 = 1 mole of CO left over)