Limiting Reactants - Chemistry

1. Fundamental Concepts

Definition: The limiting reactant is the substance that is totally consumed when the chemical reaction is complete. It "limits" the amount of product formed because the reaction cannot proceed further without it.

Excess Reactant: The substance(s) that are not used up completely in the reaction. There will be some quantity of this reactant left over after the reaction stops.

2. Key Concepts

Theoretical Yield: The maximum amount of product that can be produced is determined solely by the amount of the limiting reactant.

Mole Ratio: The stoichiometric relationship between reactants found in the balanced chemical equation dictates how much of each is needed.

Reaction Termination: The reaction stops as soon as the limiting reactant is consumed, regardless of how much excess reactant remains.

3. Examples

Easy: Basic Mole Calculation

Problem: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

You have 2 moles of $N_2$ and 5 moles of $H_2$. What is the limiting reactant?

Analysis:

From the equation: 1 mole $N_2$ needs 3 moles $H_2$.

For 2 moles $N_2$, you need $2 \times 3 = 6$ moles $H_2$.

You only have 5 moles $H_2$.

Answer: $H_2$ is the limiting reactant.

Medium:

Reaction: $4Fe(s) + 3O_2(g)→ 2Fe_2O_3(s)$

Given: 5 moles of Fe and 4 moles of O₂

Solution:

From the equation, 4 moles of Fe react with 3 moles of $O_2$.

We need 3 moles of $O_2$ for every 4 moles of Fe.

We have 5 moles of Fe, which require 3.75 moles of $O_2$.

Since we have only 4 moles of $O_2$, $O_2$ is in excess, and Fe is the limiting reactant.

Theoretical yield: 5 moles of Fe will produce 2.5 moles of $Fe_2O_3$.

Hard: Mass to Mass with Excess Calculation

Problem: $2Al(s) + 3Cl_2(g) \rightarrow 2AlCl_3(s)$

If 10.0g of Aluminum reacts with 35.0g of Chlorine gas, what is the limiting reactant? How many grams of excess reactant remain?

Analysis:

1. Convert to Moles:

$Al: 10.0g / 26.98g/mol \approx 0.371 mol$

$Cl_2: 35.0g / 70.90g/mol \approx 0.494 mol$

2. Determine Limiting:

Ratio required: 2 mol $Al$ : 3 mol $Cl_2$ (or 1 : 1.5)

If $Al$ is limiting (0.371), need $0.371 \times 1.5 = 0.5565 mol Cl_2$. We only have 0.494. Not enough $Cl_2$.

If $Cl_2$ is limiting (0.494), need $0.494 \times (2/3) \approx 0.330 mol Al$. We have 0.371. Enough $Al$.

3. Conclusion: $Cl_2$ is limiting.

4. Excess Remaining:

$Al$ used = $0.330 mol$.

$Al$ remaining = $0.371 - 0.330 = 0.041 mol$.

Mass remaining = $0.041 mol \times 26.98g/mol \approx 1.11g$.

4. Problem-Solving Techniques

There are two main methods used in US Chemistry curricula:

Method 1: The "Mole Ratio" Method
1. Balance the equation.

2. Convert all given masses to moles.

3. Pick one reactant (A) and calculate how much of the other reactant (B) is needed to react completely with A using the mole ratio.

4. Compare:

If Needed B > Available B $\rightarrow$ B is Limiting.

If Needed B < Available B $\rightarrow$ A is Limiting.

Method 2: The "Theoretical Yield" Method

1. Balance the equation.

2. Convert all given masses to moles.

3. Calculate the amount of product produced if Reactant A is completely used up.

4. Calculate the amount of product produced if Reactant B is completely used up.

5. The reactant that produces the LESSER amount of product is the Limiting Reactant.