Notes on Balancing Chemical Equations and Calculations Based on Chemical Reactions

Chapter Goals

In this chapter, we will focus on two primary goals related to chemical reactions:

1. Balancing chemical equations
2. Calculations based on chemical equations

Introduction to Chemical Equations

Chemical equations serve as symbolic representations of chemical reactions. They illustrate the following key components:

1. Reactants: These are the substances that undergo a chemical change, located on the left side of the equation.

2. Products: These are the new substances formed as a result of the chemical reaction, represented on the right side of the equation.

3. Coefficients: The numbers placed in front of the compounds to indicate the number of molecules or formula units of each reactant and product necessary to balance the equation.

   For example, the equation
   (3 ext{ Fe}2 ext{O}3 + 3 ext{CO}
   ightarrow 2 ext{Fe} + 3 ext{CO}_2)
   demonstrates these elements clearly.

Balancing Chemical Equations

To maintain the integrity of chemical equations, it is crucial that they are balanced with the smallest possible whole-number coefficients. A balanced chemical equation must have an equal number of each type of atom on both sides of the equation. This is in accordance with the Law of Conservation of Matter, which states that matter cannot be created or destroyed in an ordinary chemical reaction.

For example, the combustion reaction of propane can be expressed as:
( ext{C}3 ext{H}8 + 5 ext{O}2 ightarrow 3 ext{CO}2 + 4 ext{H}_2 ext{O}).

Practical Application of the Law of Conservation of Matter

The law can also be illustrated through the combustion of heptane (C$7$H${16}$):
( ext{C}7 ext{H}{16} + 11 ext{O}2 ightarrow 7 ext{CO}2 + 8 ext{H}_2 ext{O}). Balancing equations is a skill developed through practice, so it is advised to work through suggested problems from the chapter to gain mastery.

Calculating Based on Chemical Equations

Chemical calculations can involve working in different units such as moles, formula units, and mass (e.g., grams, kilograms). Here is an example to illustrate these applications:

• Example 1: How many CO molecules are required to react with 25 formula units of $ ext{Fe}2 ext{O}3$? Using the balanced equation:
  (9 ext{Fe}2 ext{O}3 + 3 ext{CO}
  ightarrow 2 ext{Fe} + 3 ext{CO}_2), we can determine the necessary number of CO molecules.

Mole Ratios and Calculations

From the equation, we derive the ratio of Fe$2$O$3$ to CO, which informs how many CO molecules will react with the given formula units. This ratio will dictate how we set up calculations moving forward.

For instance, using mole ratios allows us to approach the question: How many iron atoms can we produce from 2.50 x 10^5 formula units of $ ext{Fe}2 ext{O}3$?

The calculations will follow a systematic use of mole ratios derived from the original balanced equation:

• Example 2:
  rac{2 ext{Fe} ext{(produced)}}{1 ext{Fe}2 ext{O}3} ext{, where } ? ext{Fe atoms} = rac{2.50 imes 10^5 ext{ formula units}}{1 ext{ formula unit}} imes 2 ext{ Fe atoms}

Further Examples and Practice Problems

Each example builds upon previous concepts, showing various mathematical applications when calculating masses or atoms in reactions:

• Example 3 asks how much CO is needed to react with 146 g of $ ext{Fe}2 ext{O}3$, while Example 4 focuses on the mass of carbon dioxide produced from a given number of moles.

In example calculations, ensure to use proper molar masses; for instance, the molar mass of $ ext{Fe}2 ext{O}3$ is 159.7 g.

Conclusion

Understanding how to balance chemical equations and perform calculations based on them is essential in the study of chemistry. Mastery through practice is encouraged, paving the way for successful problem-solving in chemical reactions.