Question

Write a balanced equation for each of the following decomposition reactions: (a) Cobalt(III) carbonate decomposes with heat to give solid cobalt(III) oxide, and carbon dioxide gas. (b) Tin(IV) carbonate decomposes with heat to give solid tin(IV) oxide, and carbon dioxide gas.

Short answer

(a) \( \text{Co}_2(\text{CO}_3)_3 \rightarrow \text{Co}_2\text{O}_3 + 3\text{CO}_2 \); (b) \( \text{Sn(CO}_3)_2 \rightarrow \text{SnO}_2 + 2\text{CO}_2 \).

Step-by-step solution

Understanding the Decomposition Reaction

Decomposition reactions involve breaking down a compound into simpler substances. For cobalt(III) carbonate and tin(IV) carbonate, they will decompose into their respective metal oxides and carbon dioxide gas.

Identifying the Products for Cobalt(III) Carbonate

When cobalt(III) carbonate \( \text{Co}_2(\text{CO}_3)_3 \) decomposes, it produces cobalt(III) oxide \( \text{Co}_2\text{O}_3 \) and carbon dioxide gas \( \text{CO}_2 \).

Writing the Balanced Equation for Cobalt(III) Carbonate

The equation begins with the formula: \( \text{Co}_2(\text{CO}_3)_3 (s) \rightarrow \text{Co}_2\text{O}_3 (s) + 3\text{CO}_2 (g) \). Each side has 2 cobalt atoms, 3 carbon atoms, and 9 oxygen atoms, making it balanced.

Identifying the Products for Tin(IV) Carbonate

Tin(IV) carbonate \( \text{Sn(CO}_3)_2 \) decomposes into tin(IV) oxide \( \text{SnO}_2 \) and carbon dioxide gas \( \text{CO}_2 \).

Writing the Balanced Equation for Tin(IV) Carbonate

Write the equation as: \( \text{Sn(CO}_3)_2 (s) \rightarrow \text{SnO}_2 (s) + 2\text{CO}_2 (g) \). This ensures both sides have 1 tin atom, 2 carbon atoms, and 6 oxygen atoms, confirming the equation is balanced.

Key concepts

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one complex molecule breaks down into simpler ones. This process often involves the application of heat, which helps to break the chemical bonds within the compound. In the case of cobalt(III) carbonate and tin(IV) carbonate, when these compounds are heated, they decompose into metal oxides and carbon dioxide gas.
Decomposition reactions have a general formula:

• AB → A + B

Where compound AB breaks down into substances A and B. Understanding this concept is key to predicting the outcome of many chemical processes. Decomposition reactions aren't just limited to laboratory settings but occur naturally. For instance, the breaking down of limestone (calcium carbonate) to calcium oxide and carbon dioxide, similar to how cobalt(III) and tin(IV) carbonate decompose.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures that the same number of each type of atom appears on both sides of the equation. This reflects the law of conservation of mass. In our exercises, when writing the balanced equations for the decomposition of cobalt(III) carbonate and tin(IV) carbonate, it is essential to count the number of each type of atom on both sides.
For example, in the case of cobalt(III) carbonate:

• The equation starts with: \( \text{Co}_2(\text{CO}_3)_3 \rightarrow \text{Co}_2\text{O}_3 + 3\text{CO}_2 \)
• Here, we see 2 cobalt atoms, 3 carbon atoms, and 9 oxygen atoms on each side.

This equality confirms that the equation is balanced. When tackling such problems, it is a good practice to:

• Write down the number of atoms of each element from both reactants and products.
• Adjust coefficients as necessary to achieve balance without altering the molecular formulas.

Being proficient in this skill allows accurate predictions of the amounts of products formed from given reactants.

Cobalt(III) Carbonate

Cobalt(III) carbonate is a chemical compound with the formula \( \text{Co}_2(\text{CO}_3)_3 \). It consists of cobalt, carbon, and oxygen. When subjected to heat, this compound decomposes into cobalt(III) oxide \( \text{Co}_2\text{O}_3 \) and carbon dioxide gas \( \text{CO}_2 \).
The equation for this decomposition is:

• \( \text{Co}_2(\text{CO}_3)_3 (s) \rightarrow \text{Co}_2\text{O}_3 (s) + 3\text{CO}_2 (g) \)

Cobalt(III) carbonate itself can be challenging to handle due to its varying oxidation states and is used in the preparation of other cobalt compounds. Decomposition reactions of metal carbonates like this one illustrate how compounds can be systematically broken down into simpler substances, showcasing a key concept in chemistry.

Tin(IV) Carbonate

Tin(IV) carbonate, denoted as \( \text{Sn(CO}_3)_2 \), serves as another illustrative example of a metal carbonate undergoing decomposition. When heated, tin(IV) carbonate decomposes into tin(IV) oxide \( \text{SnO}_2 \) and carbon dioxide gas \( \text{CO}_2 \).
Here is the balanced reaction for its decomposition:

• \( \text{Sn(CO}_3)_2 (s) \rightarrow \text{SnO}_2 (s) + 2\text{CO}_2 (g) \)

In this process, one tin atom from the starting compound remains as part of tin(IV) oxide, a solid. This balance is essential to ensure that no atoms are lost or gained in the chemical transformation. Tin compounds like tin(IV) carbonate are valuable in various industrial applications, often exploited for their properties in semiconductor technologies and coatings.