Problem 87 - Chemistry Matter and Change

Question

Write chemical equations for each of the following decomposition reactions. One or more products might be identified. $$\begin{array}{l}{\text { a. magnesium bromide } \rightarrow} \\ {\text { b. cobalt(II) oxide } \rightarrow} \\ {\text { c. titanium(IV) hydroxide } titanium (IV) oxide + water \rightarrow} \\ {\text { d. barium carbonate } \rightarrow \text { barium oxide }+\text { carbon dioxide }}\end{array}$$

Step-by-Step Solution

Answer

Each decomposition breaks a compound into simpler substances: a. \( MgBr_2 \rightarrow \text{Mg} + \text{Br}_2 \) b. \( 2 CoO \rightarrow 2 Co + O_2 \) c. \( Ti(OH)_4 \rightarrow \text{TiO}_2 + 2 \text{H}_2\text{O} \) d. \( BaCO_3 \rightarrow \text{BaO} + \text{CO}_2 \).

Step 1: Understand the Reaction Type

The problem asks us to write chemical equations for decomposition reactions. Decomposition reactions involve a single compound breaking down into two or more simpler substances. For instance, \( AB \rightarrow A + B \).

Step 2: Decompose Magnesium Bromide

Magnesium bromide, \( MgBr_2 \), will decompose into its elements: magnesium \( Mg \) and bromine \( Br_2 \). The reaction is:\[ \text{MgBr}_2 \rightarrow \text{Mg} + \text{Br}_2 \]

Step 3: Decompose Cobalt(II) Oxide

Cobalt(II) oxide, \( CoO \), decomposes into cobalt and oxygen. The reaction is:\[ \text{2 CoO} \rightarrow \text{2 Co} + \text{O}_2 \]

Step 4: Represent Decomposition of Titanium(IV) Hydroxide

Titanium(IV) hydroxide, \( Ti(OH)_4 \), decomposes into titanium(IV) oxide and water, as stated in the problem. The equation is:\[ \text{Ti(OH)}_4 \rightarrow \text{TiO}_2 + 2 \text{H}_2\text{O} \]

Step 5: Decompose Barium Carbonate

Barium carbonate, \( BaCO_3 \), decomposes into barium oxide and carbon dioxide, as indicated. The balanced equation is:\[ \text{BaCO}_3 \rightarrow \text{BaO} + \text{CO}_2 \]

Key Concepts

Chemical Equations

Chemical equations are like sentences that describe what happens in a chemical reaction. They show the reactants (substances you start with) and the products (new substances formed). In a decomposition reaction, a single compound breaks down into two or more simpler substances. Think of it like taking apart a Lego model into individual pieces. A general formula could be written as \( AB \rightarrow A + B \). Here, compounds are broken down into their elemental forms or simpler molecules.

When writing chemical equations, it's important to balance them. This means the number of atoms for each element should be the same on both sides of the equation. Balancing ensures that the law of conservation of mass is upheld, which states that mass in a closed system is constant. A balanced equation doesn't just help chemists understand who is changing to who, but it also shows the exact amounts needed to make it happen.

Magnesium Bromide

Magnesium bromide is a compound made from magnesium and bromine, represented as \( MgBr_2 \). It consists of one magnesium ion \( Mg^{2+} \) and two bromide ions \( Br^- \). In a decomposition reaction, magnesium bromide breaks down into magnesium and bromine. It can be visualized as:

• Original: \( MgBr_2 \)
• Decomposition: \( Mg + Br_2 \)

This shows that magnesium bromide dissociates into its chemical elements, magnesium metal and diatomic bromine gas. When practicing with these reactions, ensure to maintain the correct stoichiometry to reflect this balance.

Cobalt(II) Oxide

Cobalt(II) oxide is a simple binary compound with the formula \( CoO \). It consists of cobalt in the +2 oxidation state and oxygen. Decomposition of cobalt(II) oxide involves the separation into cobalt metal and diatomic oxygen. This reaction can be expressed as:

• Original: \( 2 CoO \)
• Decomposition: \( 2 Co + O_2 \)

The balanced chemical equation reveals that it takes two units of cobalt(II) oxide to get two atoms of cobalt and one molecule of oxygen, respecting the necessity for equal amounts of cobalt and oxygen in the product.

Titanium(IV) Hydroxide

Titanium(IV) hydroxide, \( Ti(OH)_4 \), is a compound that decomposes into titanium(IV) oxide (\( TiO_2 \)) and water (\( H_2O \)). The process looks like this:

• Original: \( Ti(OH)_4 \)
• Decomposition: \( TiO_2 + 2 H_2O \)

When balanced, this equation indicates that one molecule of titanium(IV) hydroxide breaks down into one molecule of titanium(IV) oxide and two molecules of water. This type of decomposition often occurs upon heating or other catalytic actions. It's a useful reaction for producing titanium dioxide, a common material used in sunscreens and paints.

Barium Carbonate

Barium carbonate, \( BaCO_3 \), decomposes primarily into barium oxide \( BaO \) and carbon dioxide \( CO_2 \). This reaction can be expressed effectively as:

• Original: \( BaCO_3 \)
• Decomposition: \( BaO + CO_2 \)

The balanced chemical equation indicates a one-to-one conversion, where one molecule of barium carbonate yields one molecule of barium oxide and one molecule of carbon dioxide gas. This reaction is often facilitated by heating and is a foundational process in preparing barium compounds. Barium oxide formed is a crucial precursor for various industrial processes, while carbon dioxide is a common byproduct in many chemical reactions.