CH 4-3: Chemical Reactions and Stoichiometry: Mass Relationships and Limiting Reagents

Chapter 4: Chemical Reactions and Stoichiometry

Introduction

This chapter focuses on the quantitative aspects of chemical reactions, specifically mass relationships and the concept of limiting reagents. Understanding these topics is essential for predicting the amounts of products and reactants in chemical processes.

Mass Relationships and Stoichiometry

Stoichiometry and Balanced Equations

Stoichiometry is the calculation of reactants and products in chemical reactions using balanced chemical equations. A balanced equation provides the quantitative relationship between the amounts of reactants and products.

• Balanced Equation: Shows the relative number of moles of each substance involved in the reaction.

• Stoichiometric Coefficients: The numbers in front of chemical formulas in a balanced equation, indicating the ratio in which substances react and are produced.

Example: Combustion of Methane

Reactants	Products
CH4 (g) + 2 O2 (g)	CO2 (g) + 2 H2O (g)
1 molecule + 2 molecules	1 molecule + 2 molecules
1 mol + 2 mol	1 mol + 2 mol
16 g + 2(32 g)	44 g + 2(18 g)
80 g reactants	80 g products

Additional info: The table above demonstrates the law of conservation of mass, as the total mass of reactants equals the total mass of products.

Steps in Stoichiometric Calculations

1. Convert grams of substance A to moles of substance A.

2. Use the stoichiometric ratio from the balanced equation to convert moles of A to moles of B.

3. Convert moles of B to grams of B, if required.

One-step method:

• Grams of A → Moles of A → Moles of B → Grams of B

Worked Example: Stoichiometry Calculation

Reaction: 5KI + KIO3 + 6HNO3 → 6KNO3 + 3I2 + 3H2O

Problem: What mass of I2 is produced if 13.10 g KI is reacted with excess KIO3 and HNO3?

• Molar mass (M) KI = 166.0 g/mol

• Molar mass (M) I2 = 253.8 g/mol

Solution:

• Convert KI to moles: $13.10\,\text{g} \times \frac{1\,\text{mol}}{166.0\,\text{g}} = 0.07892\,\text{mol}$

• Use stoichiometric ratio: $0.07892\,\text{mol KI} \times \frac{3\,\text{mol I}_2}{5\,\text{mol KI}} = 0.04735\,\text{mol I}_2$

• Convert I2 to grams: $0.04735\,\text{mol} \times 253.8\,\text{g/mol} = 12.02\,\text{g}$

Limiting Reagents

Definition and Identification

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

• The moles of product formed and the amount of other reactants consumed are always determined by the starting moles of the limiting reagent.

Steps to Identify the Limiting Reagent

1. Convert all given reactant masses to moles.

2. Divide the number of moles of each reactant by its coefficient in the balanced equation (mole-coefficient ratio).

3. The reactant with the smallest mole-coefficient ratio is the limiting reagent.

Additional info: This is one of several techniques for limiting reagent problems. Some textbooks may use a different approach, but the principle remains the same.

Worked Example: Limiting Reagent Calculation

Reaction: 2Ca3(PO4)2 + 6SiO2 + 10C → 6CaSiO3 + 10CO + P4

• Given: 6.00 g Ca3(PO4)2, 4.00 g SiO2, 1.00 g C

• M(Ca3(PO4)2) = 310 g/mol, M(SiO2) = 60.1 g/mol, M(C) = 12.01 g/mol

Substance	Coefficient	Moles	Mole-Coefficient Ratio
Ca3(PO4)2	2	0.0194	0.0097
SiO2	6	0.0666	0.0111
C	5	0.0833	0.0167

• Limiting reagent: Ca3(PO4)2 (smallest ratio)

• Maximum mass of P4 produced: $0.00970\,\text{mol} \times 123.9\,\text{g/mol} = 1.20\,\text{g}$

Worked Example: Multi-Step Limiting Reagent Problem

Step 1: 4HCl + MnO2 → MnCl2 + 2H2O + Cl2

Step 2: P4 + 6Cl2 → 4PCl3

• Given: 10.0 g HCl, 6.08 g MnO2

• Find: Maximum mass of PCl3 that can be prepared

Reagent	Moles	Coefficient	Mole-Coefficient Ratio
HCl	0.274	4	0.0685
MnO2	0.0699	1	0.0699

• Limiting reagent: HCl (smallest ratio)

• Linking the two equations via stoichiometry:

$0.274\,\text{mol HCl} \times \frac{1\,\text{mol Cl}_2}{4\,\text{mol HCl}} \times \frac{4\,\text{mol PCl}_3}{6\,\text{mol Cl}_2} = 0.0457\,\text{mol PCl}_3$

Mass of PCl3: $0.0457\,\text{mol} \times 137.4\,\text{g/mol} = 6.28\,\text{g}$

Key Terms and Concepts

• Stoichiometry: The calculation of reactants and products in chemical reactions.

• Limiting Reagent: The reactant that is completely consumed first, limiting the amount of product formed.

• Excess Reagent: Reactants that are not completely consumed in the reaction.

• Mole Ratio: The ratio of moles of one substance to moles of another as indicated by the coefficients in a balanced equation.

Summary Table: Steps in Stoichiometric Calculations

Step	Description
1	Convert mass of reactant to moles
2	Use mole ratio from balanced equation to find moles of product
3	Convert moles of product to mass (if required)