 # 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)

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## 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)

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## 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) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

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

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## 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)

Footnotes:

(1) A reaction that goes to completion is a spontaneous, irreversible chemical reaction.
The arrow used to indicate this in the balanced chemical equation is →
The system is not at equilibrium so the equilibrium arrow, ⇋, is not used.

(2) Concentration in mol L-1 (molarity) is not the only way to measure concentration of solutions. But for the sake of this introductory tutorial we will only be considering molarity of solutions in our examples.

(3) (i) You can also use Gay-Lussac's Law of Combining Gas Volumes if all the reactants and products are gases.
(ii) To calculate the amount of ideal gas in moles for other temperatures and pressures (not STP or SLC) use the Ideal Gas Equation.