Question

Balance each of the following equations that describe decomposition reactions. a. \(\operatorname{CaSO}_{4}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{SO}_{3}(g)\) b. \(\mathrm{Li}_{2} \mathrm{CO}_{3}(s) \rightarrow \mathrm{Li}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)\) c. \(\operatorname{LiHCO}_{3}(s) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)\) d. \(C_{6} H_{6}(l) \rightarrow C(s)+H_{2}(g)\) e. \(\operatorname{PBr}_{3}(l) \rightarrow \mathrm{P}_{4}(s)+\mathrm{Br}_{2}(l)\)

Short answer

a. \(\operatorname{CaSO}_{4}(s) \rightarrow \mathrm{CaO}(s)+2\mathrm{SO}_{3}(g)\) b. \(\mathrm{Li}_{2} \mathrm{CO}_{3}(s) \rightarrow \mathrm{Li}_{2}\mathrm{O}(s)+\mathrm{CO}_2(g)\) c. \(2\operatorname{LiHCO}_{3}(s) \rightarrow \mathrm{Li}_{2}\mathrm{CO}_{3}(s)+\mathrm{H}_{2}\mathrm{O}(g)+\mathrm{CO}_{2}(g)\) d. \(C_{6}H_{6}(l) \rightarrow 6C(s) + 3H_{2}(g)\) e. \(4\operatorname{PBr}_{3}(l) \rightarrow \mathrm{P}_{4}(s) + 6\mathrm{Br}_{2}(l)\)

Step-by-step solution

a. Balancing the equation for \(\operatorname{CaSO}_{4}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{SO}_{3}(g)\)

Since we have 1 \(\mathrm{Ca}\) atom, 1 \(\mathrm{S}\) atom, and 4 \(\mathrm{O}\) atoms on the left side of the equation, and only 1 \(\mathrm{S}\) atom and 3 \(\mathrm{O}\) atoms on the right side, we need to add a coefficient of 2 in front of \(\mathrm{SO}_{3}(g)\) in order to balance the number of oxygen atoms. The balanced equation is: \(\operatorname{CaSO}_{4}(s) \rightarrow \mathrm{CaO}(s)+2\mathrm{SO}_{3}(g)\)

b. Balancing the equation for \(\mathrm{Li}_{2} \mathrm{CO}_{3}(s) \rightarrow \mathrm{Li}_{2}\mathrm{O}(s)+\mathrm{CO}_2(g)\)

We have 2 \(\mathrm{Li}\) atoms, 1 \(\mathrm{C}\) atom, and 3 \(\mathrm{O}\) atoms on the left side, and 2 \(\mathrm{Li}\) atoms, 1 \(\mathrm{C}\) atom, and 3 \(\mathrm{O}\) atoms on the right side. This equation is correctly balanced as is.

c. Balancing the equation for \(\operatorname{LiHCO}_{3}(s) \rightarrow \mathrm{Li}_{2}\mathrm{CO}_{3}(s)+\mathrm{H}_{2}\mathrm{O}(g)+\mathrm{CO}_{2}(g)\)

We will have to multiply LiHCO3 by 2 to get equal lithium atoms on both sides and equal hydrogen atoms. The balanced equation is: \(2\operatorname{LiHCO}_{3}(s) \rightarrow \mathrm{Li}_{2}\mathrm{CO}_{3}(s)+\mathrm{H}_{2}\mathrm{O}(g)+\mathrm{CO}_{2}(g)\)

d. Balancing the equation for \(C_{6}H_{6}(l) \rightarrow C(s)+H_{2}(g)\)

We have 6 carbon atoms and 6 hydrogen atoms on the left inside the C6H6 compound. To balance it we should multiply C by 6 on the right side (6C) and multiply H2 by 3 (3H2) to have the same number of hydrogen atoms. The balanced equation is: \(C_{6}H_{6}(l) \rightarrow 6C(s) + 3H_{2}(g)\).

e. Balancing the equation for \(\operatorname{PBr}_{3}(l) \rightarrow \mathrm{P}_{4}(s)+\mathrm{Br}_{2}(l)\)

Here we have 1 \(\mathrm{P}\) atom and 3 \(\mathrm{Br}\) atoms on the left side and 4 \(\mathrm{P}\) atoms and 2 \(\mathrm{Br}\) atoms on the right side. To balance this equation, we need to multiply \(\operatorname{PBr}_{3}\) by 4, to get equal phosphorus atoms on both sides. In addition, we will have to multiply \(\mathrm{Br}_2(l)\) by 6 to have 12 bromine atoms on both sides. The balanced equation is: \(4\operatorname{PBr}_{3}(l) \rightarrow \mathrm{P}_{4}(s) + 6\mathrm{Br}_{2}(l)\).

Key concepts

Decomposition Reactions

Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. This reaction is characterized by the breakdown of a reactant into multiple products.
These reactions often require an input of energy, such as heat, light, or electricity, to proceed. A classic example of a decomposition reaction is the breakdown of water (\(2\mathrm{H}_2\mathrm{O}(l) \rightarrow 2\mathrm{H}_2(g) + \mathrm{O}_2(g)\)) when electricity is applied, known as electrolysis.

In the exercise above, each equation begins with a single compound that decomposes into smaller molecules or elements. Understanding decomposition reactions is crucial because it helps in predicting the products of a variety of reactions and is widely used in industrial processes.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through the breaking and forming of chemical bonds. They are fundamental to understanding how substances interact and change.
Every chemical reaction can be represented by a chemical equation that depicts both the reactants and the products, providing insight into the transformation that takes place.

Consider the decomposition of calcium sulfate described in the exercise. The chemical reaction is represented as:

• Reactant: \(\mathrm{CaSO}_4(s)\)
• Products: \(\mathrm{CaO}(s) + \mathrm{SO}_3(g)\)

Chemical reactions are often categorized into different types, including synthesis, decomposition, single replacement, double replacement, and combustion reactions, each defined by the change in the makeup of the molecules involved.

Balancing Equations

Balancing chemical equations is an essential skill in chemistry to ensure the law of conservation of mass is obeyed. This law states that matter cannot be created or destroyed in a chemical reaction, meaning the number of atoms of each element must be the same on both sides of the equation.
Balancing equations involves adjusting coefficients to make sure the atoms of each element are equal in the reactants and products. For example, in the given exercise:

• In \(\mathrm{C}_{6} \mathrm{H}_{6}(l) \rightarrow \mathrm{C}(s) + \mathrm{H}_{2}(g)\), we multiply \(\mathrm{C}\) by 6 and \(\mathrm{H}_{2}\) by 3 to balance the equation.
• Checking each side after adjustments ensures the equation is balanced.

Efficiently balancing reactions helps in understanding the exact proportions needed for reactions to occur correctly.

Stoichiometry

Stoichiometry is the area of chemistry that pertains to the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction.
Using stoichiometry involves using coefficients from a balanced chemical equation to calculate the relative quantities. Take this balanced equation from our exercise:

• \(4\mathrm{PBr}_{3}(l) \rightarrow \mathrm{P}_{4}(s) + 6\mathrm{Br}_{2}(l)\)
• The coefficients (such as 4 for \(\mathrm{PBr}_{3}\)) help in determining how much of each substance is consumed or produced.

Stoichiometry is crucial when dealing with chemical reactions, where specific quantities of a reactant are necessary to ensure complete reactions, minimizing waste and ensuring safety in industrial processes.