Problem 85 - Chemistry Matter and Change

Question

Write chemical equations for each of the following synthesis reactions. $$\begin{array}{l}{\text { a. boron }+\text { fluorine } \rightarrow} \\\ {\text { b. germanium }+\text { sulfur } \rightarrow} \\ {\text { c. zirconium }+\text { nitrogen } \rightarrow} \\ {\text { d. tetraphosphorus decoxide + water } \rightarrow \text { phosphoric acid }}\end{array}$$

Step-by-Step Solution

Answer

Boron fluoride: \(\text{2B} + \text{3F}_2 \rightarrow \text{2BF}_3\). Germanium sulfide: \(\text{Ge} + \text{2S} \rightarrow \text{GeS}_2\). Zirconium nitride: \(\text{3Zr} + \text{2N}_2 \rightarrow \text{Zr}_3\text{N}_4\). Phosphoric acid: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\).

Step 1: Write the Reactants

Identify the reactants given in each synthesis reaction.- For reaction a: The reactants are Boron (B) and Fluorine (F).- For reaction b: The reactants are Germanium (Ge) and Sulfur (S).- For reaction c: The reactants are Zirconium (Zr) and Nitrogen (N).- For reaction d: The reactants are Tetraphosphorus decoxide \((\text{P}_4\text{O}_{10})\) and Water \((\text{H}_2\text{O})\).

Step 2: Determine the Products

Figure out the likely products for each reaction based on common chemical combinations.- For reaction a: Boron and Fluorine form Boron Trifluoride \((\text{BF}_3)\).- For reaction b: Germanium and Sulfur form Germanium Sulfide \((\text{GeS}_2)\).- For reaction c: Zirconium and Nitrogen form Zirconium Nitride \((\text{Zr}_3\text{N}_4)\).- For reaction d: Tetraphosphorus decoxide and water form Phosphoric Acid \((\text{H}_3\text{PO}_4)\).

Step 3: Write Balanced Chemical Equations

Write out and balance the chemical equations for each reactant and product pair.- Reaction a: \(\text{2B} + \text{3F}_2 \rightarrow \text{2BF}_3\)- Reaction b: \(\text{Ge} + \text{2S} \rightarrow \text{GeS}_2\)- Reaction c: \(\text{3Zr} + \text{2N}_2 \rightarrow \text{Zr}_3\text{N}_4\)- Reaction d: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\)

Key Concepts

Understanding Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction. It shows the substances that are involved in the reaction, known as reactants, and those that are produced, known as products. Chemical equations help predict the outcomes of a reaction and provide a simple way to depict complex reactions.

In a chemical equation, the reactants are written on the left side, and the products are written on the right side. An arrow separates these two parts, indicating the direction of the reaction. To read a chemical equation correctly, it's important to understand the symbols and numbers, which provide information about the quantities and types of atoms involved in the reactions. For example, the equation \( \text{2B} + \text{3F}_2 \rightarrow \text{2BF}_3 \) indicates that 2 atoms of boron react with 3 molecules of fluorine to form 2 molecules of boron trifluoride.

Mastering chemical equations involves learning to identify all components of the reactions, understanding the symbols, and being able to write and interpret the equations accurately.

Reactants and Products

Reactants and products are the fundamental components of any chemical equation. Reactants are the starting substances, while products are the new substances formed as a result of the chemical reaction. Understanding the difference between reactants and products is essential for describing chemical reactions accurately.

In synthesis reactions, like those provided in the exercise, multiple reactants combine to form a single product. For example, in the reaction of boron and fluorine, boron and fluorine (the reactants) combine to form boron trifluoride (the product).

Here's a simple way to remember their roles:

• **Reactants:** The substances you start with before the reaction.
• **Products:** The substances you end up with after the reaction.

In our example, the reactants are Boron \((\text{B})\) and Fluorine \((\text{F}_2)\), and the product is Boron Trifluoride \((\text{BF}_3)\). Get familiar with identifying and differentiating between these components to excel in chemistry.

Balancing Equations

Balancing chemical equations is a crucial skill in chemistry. A balanced equation ensures that the same number of each type of atom is present on both sides of the equation, thus obeying the Law of Conservation of Mass. This law states that matter cannot be created or destroyed in a chemical reaction.

To balance equations, follow these steps:

• Write the unbalanced equation with the correct formulas for each reactant and product.
• Count the number of atoms of each element on both sides of the equation.
• Add coefficients in front of the formulas to balance the atoms, without changing the actual formulas of the substances.

For instance, in the reaction \( \text{2B} + \text{3F}_2 \rightarrow \text{2BF}_3 \), coefficients are added to ensure 2 boron atoms and 6 fluorine atoms are on each side. This balancing not only complies with chemical laws but also gives accurate representations for practical applications in lab settings and industry.

Exploring Inorganic Chemistry

Inorganic chemistry deals with the study of compounds that are not covered by organic chemistry, which primarily focuses on carbon-containing compounds. Inorganic chemistry encompasses a vast array of substances including metals, minerals, and organometallic compounds.

Synthesis reactions, like those involving boron and fluorine or zirconium and nitrogen, are typical in inorganic chemistry. They often involve elements from the periodic table combining in various ways to create new compounds. Understanding inorganic reactions is crucial because they apply to many real-world applications such as materials engineering, catalysis, and environmental science.

Inorganic chemistry broadly considers:

• **Elements and compounds:** Metals, non-metals, and their oxides, sulfides, and halides.
• **Reactions:** Synthesis, decomposition, single displacement, and double displacement reactions.
• **Applications:** Creating catalysts, developing pharmaceuticals, and designing advanced materials.

This field of chemistry is vast and offers exposure to many types of reactions, including those seen in textbook exercises, helping us understand the foundational principles before exploring more specialized areas.