Limiting Reagents

Overview

The limiting reagent (or limiting reactant) is the reactant that is completely consumed first in a chemical reaction, determining the maximum amount of product that can be formed. The other reactant(s) are in excess.

Key Concepts

Limiting Reagent

• Completely consumed in the reaction
• Determines the amount of product formed
• Limits the reaction from continuing

Excess Reagent

• Not completely consumed
• Some remains after reaction
• Present in more than stoichiometric amount

Finding the Limiting Reagent

Method 1: Compare Mole Ratios

1. Convert all reactants to moles
2. Divide each by its coefficient in the balanced equation
3. The smallest value indicates the limiting reagent

Method 2: Calculate Product from Each Reactant

1. Assume each reactant is limiting
2. Calculate product amount from each
3. The one giving less product is limiting

Step-by-Step Example

Problem

Given 10.0 g of hydrogen and 80.0 g of oxygen, how much water forms?

$$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$

Solution

Step 1: Convert to moles

$$\text{mol H}_2 = \frac{10.0 \text{ g}}{2.016 \text{ g/mol}} = 4.96 \text{ mol}$$ $$\text{mol O}_2 = \frac{80.0 \text{ g}}{32.00 \text{ g/mol}} = 2.50 \text{ mol}$$

Step 2: Find limiting reagent (Method 1)

$$\text{H}_2: \frac{4.96}{2} = 2.48$$ $$\text{O}_2: \frac{2.50}{1} = 2.50$$

H₂ is limiting (smaller ratio)

Step 3: Calculate product

$$\text{mol H}_2\text{O} = 4.96 \text{ mol H}_2 \times \frac{2 \text{ mol H}_2\text{O}}{2 \text{ mol H}_2} = 4.96 \text{ mol}$$ $$\text{mass H}_2\text{O} = 4.96 \text{ mol} \times 18.02 \text{ g/mol} = 89.4 \text{ g}$$

Step 4: Calculate excess

$$\text{O}_2 \text{ used} = 4.96 \text{ mol H}_2 \times \frac{1 \text{ mol O}_2}{2 \text{ mol H}_2} = 2.48 \text{ mol}$$ $$\text{O}_2 \text{ excess} = 2.50 - 2.48 = 0.02 \text{ mol} = 0.64 \text{ g}$$

More Examples

Example 2: Synthesis of Ammonia

$$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$$

Given: 28.0 g N₂ and 10.0 g H₂

$$\text{mol N}_2 = \frac{28.0}{28.02} = 0.999 \text{ mol}$$ $$\text{mol H}_2 = \frac{10.0}{2.016} = 4.96 \text{ mol}$$

Ratio check:

$$\text{N}_2: \frac{0.999}{1} = 0.999$$ $$\text{H}_2: \frac{4.96}{3} = 1.65$$

N₂ is limiting (smaller)

$$\text{mol NH}_3 = 0.999 \times \frac{2}{1} = 2.00 \text{ mol}$$ $$\text{mass NH}_3 = 2.00 \times 17.03 = 34.1 \text{ g}$$

Example 3: Precipitation Reaction

$$2\text{AgNO}_3(aq) + \text{CaCl}_2(aq) \rightarrow 2\text{AgCl}(s) + \text{Ca(NO}_3\text{)}_2(aq)$$

Given: 0.100 mol AgNO₃ and 0.100 mol CaCl₂

Ratio check:

$$\text{AgNO}_3: \frac{0.100}{2} = 0.050$$ $$\text{CaCl}_2: \frac{0.100}{1} = 0.100$$

AgNO₃ is limiting

$$\text{mol AgCl} = 0.100 \text{ mol AgNO}_3 \times \frac{2}{2} = 0.100 \text{ mol}$$

Stoichiometric Calculations Summary

Conversion Pathway

$$\text{Given mass} \xrightarrow{\div \text{molar mass}} \text{Moles of given} \xrightarrow{\times \text{mole ratio}} \text{Moles of wanted} \xrightarrow{\times \text{molar mass}} \text{Mass of product}$$

General Formula

$$\text{mass of product} = \frac{\text{given mass}}{\text{given molar mass}} \times \frac{\text{product coefficient}}{\text{given coefficient}} \times \text{product molar mass}$$

Common Limiting Reagent Scenarios

Scenario	Limiting Reagent
A + B → C, equal moles, 1:1 ratio	Neither (stoichiometric)
A + 2B → C, equal moles	B is limiting
2A + B → C, equal moles	A is limiting
Large excess of one reactant	The other is limiting

Tips for Problem Solving

1. Always balance the equation first
2. Convert everything to moles
3. Use mole ratios, not mass ratios
4. The limiting reagent gives the smaller amount of product
5. To find excess, calculate how much was used and subtract