Limiting Reagent Calculations Revision Notes for HSC SSCE Chemistry

Limiting Reagent Calculations

Introduction to limiting reagents

When performing chemical reactions in the laboratory, you often mix specific amounts of two or more reactants together. In previous topics, you learned how to calculate the amount of one reactant needed to completely react with a given amount of another reactant. However, in real-world situations, you may have known quantities of both reactants, and you need to determine how much product will form.

In these situations, one reactant will typically be completely consumed while some of the other reactant remains unreacted. The reactant that is completely consumed during the reaction is known as the limiting reagent (or limiting reactant). This reagent is called "limiting" because it determines the maximum amount of product that can be formed. The reactant that is not completely used up is called the excess reagent.

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Think of it like baking: if you have enough ingredients to make 10 cakes except you only have 3 eggs, the eggs become your limiting ingredient. You can only make as many cakes as your egg supply allows, even if you have excess flour, sugar, and butter.

Why limiting reagents are important

Understanding limiting reagents is crucial because:

In laboratory work, chemists commonly use an excess of one reactant to ensure the other reactant is completely consumed
The amount of product formed depends entirely on the limiting reagent
You cannot base your calculations on the excess reagent, as some of it remains after the reaction is complete
Identifying the limiting reagent allows you to predict the maximum yield of product

Identifying the limiting reagent

To identify which reactant is the limiting reagent, follow this approach:

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Identification Method:

Start with one reactant - Choose either reactant as your starting point
Calculate the stoichiometric requirement - Determine how much of the other reactant is needed to completely react with your chosen reactant
Compare with available amount - Check if you have enough of the second reactant
Determine the limiting reagent - If you have insufficient amounts, then the second reactant is limiting. If you have excess, then the first reactant is limiting

The key principle is to use molar ratios from the balanced chemical equation to compare the amounts available versus the amounts required.

Step-by-step method for limiting reagent calculations

When solving limiting reagent problems, follow these steps:

Step 1: Write the balanced chemical equation for the reaction

Step 2: Calculate the number of moles of each reactant provided (if given masses, convert using $n = \frac{m}{M}$ )

Step 3: Use stoichiometric ratios to determine which reactant is limiting:

Choose one reactant and calculate how much of the other reactant is needed
Compare this calculated amount with the available amount
The reactant that runs out first is the limiting reagent

Step 4: Calculate the amount of product using only the limiting reagent

Step 5: If required, determine how much of the excess reagent remains unreacted

Worked example 1: Using moles directly

Consider a reaction between sulfuric acid and aluminium hydroxide:

$2\text{Al(OH)}_3\text{(s)} + 3\text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Al}_2(\text{SO}_4)_3\text{(aq)} + 6\text{H}_2\text{O(l)}$

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Worked Example: Identifying the Limiting Reagent Using Moles

Problem: If $6 \text{ mol}$ sulfuric acid and $5 \text{ mol}$ aluminium hydroxide are mixed, which reactant is completely used up? How many moles of which reagent remain unreacted? How many moles of aluminium sulfate are formed?

Solution:

From the balanced equation, the molar ratio is:

$\frac{\text{moles of H}_2\text{SO}_4}{\text{moles of Al(OH)}_3} = \frac{3}{2}$

To use up all $5 \text{ mol}$ of $\text{Al(OH)}_3$ , we need:

$\text{moles of H}_2\text{SO}_4 \text{ required} = \frac{3}{2} \times 5 = 7.5 \text{ mol}$

However, we only have $6 \text{ mol}$ of $\text{H}_2\text{SO}_4$ available. This is insufficient to react with all the aluminium hydroxide. Therefore, sulfuric acid is the limiting reagent.

Now calculate how much $\text{Al(OH)}_3$ will actually react with the $6 \text{ mol}$ of $\text{H}_2\text{SO}_4$ :

$\frac{\text{moles of Al(OH)}_3}{\text{moles of H}_2\text{SO}_4} = \frac{2}{3}$

$\text{moles of Al(OH)}_3 \text{ that react} = \frac{2}{3} \times 6 = 4 \text{ mol}$

This is less than the $5 \text{ mol}$ provided, so:

$\text{moles of Al(OH)}_3 \text{ left over} = 5 - 4 = 1 \text{ mol}$

To find the product formed, use the limiting reagent ( $\text{H}_2\text{SO}_4$ ):

$\frac{\text{moles of Al}_2(\text{SO}_4)_3}{\text{moles of H}_2\text{SO}_4} = \frac{1}{3}$

$\text{moles of Al}_2(\text{SO}_4)_3 \text{ formed} = \frac{1}{3} \times 6 = 2 \text{ mol}$

Answer: Sulfuric acid is completely used up, $1 \text{ mol}$ of aluminium hydroxide remains unreacted, and 2 mol of aluminium sulfate are formed.

Worked example 2: Using masses of reactants

When reactant quantities are given as masses rather than moles, you must first convert to moles before applying the same limiting reagent logic.

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Worked Example: Limiting Reagent Calculation Using Masses

Problem: $4.1 \text{ g}$ potassium iodide is dissolved in water and added to a solution containing $4.8 \text{ g}$ lead nitrate to produce a precipitate of yellow lead iodide. What mass of lead iodide is formed?

The balanced equation is:

$\text{Pb(NO}_3)_2\text{(aq)} + 2\text{KI(aq)} \rightarrow \text{PbI}_2\text{(s)} + 2\text{KNO}_3\text{(aq)}$

Solution:

Step 1: Calculate molar masses

$M(\text{KI}) = 39.10 + 126.9 = 166.0 \text{ g mol}^{-1}$

$M(\text{Pb(NO}_3)_2) = 207.2 + 2(14.00 + 3 \times 16.00) = 331.2 \text{ g mol}^{-1}$

Step 2: Convert masses to moles

$n(\text{Pb(NO}_3)_2) = \frac{4.8}{331.2} = 0.0145 \text{ mol}$

$n(\text{KI}) = \frac{4.1}{166.0} = 0.0247 \text{ mol}$

Step 3: Identify the limiting reagent

From the balanced equation, the stoichiometric ratio shows that $2 \text{ mol}$ of $\text{KI}$ reacts with $1 \text{ mol}$ of $\text{Pb(NO}_3)_2$ .

To use up all the $\text{Pb(NO}_3)_2$ :

$\text{moles of KI needed} = 2 \times 0.0145 = 0.0290 \text{ mol}$

We only have $0.0247 \text{ mol}$ of $\text{KI}$ available. This is less than required, so KI is the limiting reagent. Some lead nitrate will remain unreacted.

Step 4: Calculate product based on limiting reagent

From the equation, the molar ratio is:

$\frac{\text{moles of PbI}_2}{\text{moles of KI}} = \frac{1}{2}$

$n(\text{PbI}_2) = \frac{1}{2} \times 0.0247 = 0.0124 \text{ mol}$

Step 5: Convert moles of product to mass

$M(\text{PbI}_2) = 207.2 + 2 \times 126.9 = 461.0 \text{ g mol}^{-1}$

$m(\text{PbI}_2) = 0.0124 \times 461.0 = 5.7 \text{ g}$

Answer: 5.7 g of lead iodide is formed.

chatImportant

Key Calculation Tips:

Always start by writing the balanced chemical equation
When masses are given, convert to moles first using $n = \frac{m}{M}$
Use stoichiometric ratios from the balanced equation
The limiting reagent is the one that produces the smallest amount of product
Base all product calculations on the limiting reagent only
Remember to round your final answer to the appropriate number of significant figures

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Key Points to Remember:

The limiting reagent is the reactant that is completely consumed in a chemical reaction. It determines the maximum amount of product that can be formed.
The excess reagent is the reactant that remains after the reaction is complete. You cannot use it to calculate product amounts.
To identify the limiting reagent: Calculate how much of one reactant is needed to use up the other, then compare this with the available amount. The reagent that runs out first is limiting.
Always use the limiting reagent for product calculations. Any calculation based on the excess reagent will give an incorrect (too large) answer.
When given masses, convert to moles first using the molar mass formula: $n = \frac{m}{M}$ . Then apply stoichiometric ratios from the balanced equation.

Limiting Reagent Calculations (HSC SSCE Chemistry): Revision Notes