Limiting Reagent

The Concept of the Limiting Reagent

In any chemical reaction, the reactants are rarely present in the exact ratio demanded by the balanced chemical equation. Usually, one reactant is taken in excess, while the other is present in a smaller amount. The reaction proceeds until the reactant that is present in the smallest stoichiometric amount is completely consumed. Once this reactant is used up, the reaction stops, regardless of how much of the other reactant remains. This reactant is called the limiting reagent (or limiting reactant).

The limiting reagent determines the maximum amount of product that can be formed. Any reactant that is not completely consumed is called an excess reagent.

How to Identify the Limiting Reagent

The textbook outlines a clear, step-by-step method for identifying the limiting reagent in a given reaction. Consider a general reaction:

where $a$, $b$, $c$, and $d$ are the stoichiometric coefficients from the balanced equation.

Step 1: Calculate the number of moles of each reactant present.

If the mass of a reactant is given, use its molar mass:

If the volume and concentration of a solution are given, use:

Step 2: Determine the stoichiometric requirement.

From the balanced equation, $a$ moles of A react with $b$ moles of B. Therefore, the amount of B required to completely react with the given moles of A is:

Step 3: Compare the required amount with the available amount.

• If the available moles of B are greater than or equal to the required moles of B, then A is the limiting reagent. B is in excess.
• If the available moles of B are less than the required moles of B, then B is the limiting reagent. A is in excess.

Calculating the Amount of Product Formed

Once the limiting reagent is identified, it is used to calculate the theoretical yield of any product. The calculation is based on the mole ratio between the limiting reagent and the product, as given by the balanced chemical equation.

For the reaction $aA + bB \rightarrow cC$, if A is the limiting reagent, then:

The mass of C produced can then be found by multiplying the moles of C by its molar mass.

Worked Example

Let's apply the method step-by-step to a simple practice case of our own. (The textbook's own worked limiting-reagent case is Problem 1.5 — the ammonia synthesis — which has its own question page in this chapter.)

Problem: In a reaction, 2.0 g of hydrogen gas ($H_2$) is mixed with 16.0 g of oxygen gas ($O_2$) to form water ($H_2O$). Identify the limiting reagent and calculate the mass of water produced.

Step 1: Write the balanced chemical equation.

Step 2: Calculate the initial moles of each reactant.

• Molar mass of $H_2$ = $2 \times 1.008 \text{ g mol}^{-1} = 2.016 \text{ g mol}^{-1}$

• Moles of $H_2$ = $\frac{2.0 \text{ g}}{2.016 \text{ g mol}^{-1}} \approx 0.992 \text{ mol}$

• Molar mass of $O_2$ = $2 \times 16.00 \text{ g mol}^{-1} = 32.00 \text{ g mol}^{-1}$

• Moles of $O_2$ = $\frac{16.0 \text{ g}}{32.00 \text{ g mol}^{-1}} = 0.500 \text{ mol}$

Step 3: Identify the limiting reagent using the mole-ratio method.

From the balanced equation, 2 moles of $H_2$ react with 1 mole of $O_2$.

• Method 1 (Stoichiometric requirement):

  • Moles of $O_2$ required to react with 0.992 mol of $H_2$ = $\frac{1}{2} \times 0.992 = 0.496 \text{ mol}$
  • Available $O_2$ = 0.500 mol
  • Since the available $O_2$ (0.500 mol) is greater than the required $O_2$ (0.496 mol), $H_2$ is the limiting reagent.

• Method 2 (Dividing by coefficient):

  • For $H_2$: $\frac{0.992}{2} = 0.496$ …