Question

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

Short answer

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

Step-by-step solution

Balance the Calcium atoms

Since there is one calcium atom on each side of the equation, the calcium atoms are already balanced.

Balance the Sulfur atoms

There is one sulfur atom on each side of the equation, so the sulfur atoms are also balanced.

Balance the Oxygen atoms

There are four oxygen atoms on the left side and three on the right side of the equation. To balance the oxygen atoms, we multiply SO3 by 4/3. The balanced equation becomes: $$\mathrm{CaSO}_{4}(s) \rightarrow \mathrm{CaO}(s)+\frac{4}{3}\mathrm{SO}_{3}(g)$$ b. Balancing Li2CO3(s) → Li2O(s) + CO2(g):

Balance the Lithium atoms

There are two lithium atoms on both sides of the equation, so the lithium atoms are balanced.

Balance the Carbon atoms

There is one carbon atom on both sides of the equation, so the carbon atoms are balanced.

Balance the Oxygen atoms

There are three oxygen atoms on both sides of the equation, so the oxygen atoms are balanced. The equation is balanced overall. c. Balancing LiHCO3(s) → Li2CO3(s) + H2O(g) + CO2(g):

Balance the Lithium atoms

There is one lithium atom on the left side of the equation and two on the right side. To balance the lithium atoms, we multiply LiHCO3 by 2. The equation becomes: $$2\mathrm{LiHCO}_{3}(s) \rightarrow \mathrm{Li}_{2}\mathrm{CO}_{3}(s)+\mathrm{H}_{2}\mathrm{O}(g)+\mathrm{CO}_{2}(g)$$

Balance the Carbon atoms

There are two carbon atoms on the left side and three on the right side. To balance the carbon atoms, we multiply CO2 by 2/3. The equation becomes: $$2\mathrm{LiHCO}_{3}(s) \rightarrow \mathrm{Li}_{2}\mathrm{CO}_{3}(s)+\mathrm{H}_{2}\mathrm{O}(g)+\frac{2}{3}\mathrm{CO}_{2}(g)$$

Balance the Hydrogen atoms

Now, there are two hydrogen atoms on both sides of the equation, so the hydrogen atoms are balanced.

Balance the Oxygen atoms

Finally, there are six oxygen atoms on both sides of the equation, so the oxygen atoms are balanced. The equation is balanced overall. d. Balancing C6H6(l) → C(s) + H2(g):

Balance the Carbon atoms

There are six carbon atoms on the left side and one on the right side. To balance the carbon atoms, we multiply C by 6. The equation becomes: $$\mathrm{C}_{6}\mathrm{H}_{6}(l) \rightarrow 6\mathrm{C}(s)+\mathrm{H}_{2}(g)$$

Balance the Hydrogen atoms

There are six hydrogen atoms on the left side and two on the right side. To balance the hydrogen atoms, we multiply H2 by 3. The balanced equation becomes: $$\mathrm{C}_{6}\mathrm{H}_{6}(l) \rightarrow 6\mathrm{C}(s)+3\mathrm{H}_{2}(g)$$ e. Balancing PBr3(l) → P4(s) + Br2(l):

Balance the Phosphorus atoms

There is one phosphorus atom on the left side and four on the right side. To balance the phosphorus atoms, we multiply PBr3 by 4. The equation becomes: $$4\mathrm{PBr}_{3}(l) \rightarrow \mathrm{P}_{4}(s)+\mathrm{Br}_{2}(l)$$

Balance the Bromine atoms

There are twelve bromine atoms on the left side and two on the right side. To balance the bromine atoms, we multiply Br2 by 6. The balanced equation becomes: $$4\mathrm{PBr}_{3}(l) \rightarrow \mathrm{P}_{4}(s)+6\mathrm{Br}_{2}(l)$$

Key concepts

Chemical Equation Balancing

Balancing chemical equations is an essential process in chemistry that ensures the law of conservation of mass is observed. In a balanced chemical equation, the number of atoms of each element is the same on both sides of the reaction. This principle applies to all types of chemical reactions, including decomposition reactions.

For example, when balancing the decomposition reaction \(\mathrm{CaSO}_{4}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{SO}_{3}(g)\), it is crucial to ensure that the number of calcium, sulfur, and oxygen atoms remains unchanged from reactants to products. Steps for balancing involve adjusting coefficients, the number preceding compounds, to equalize the number of atoms for each element across the equation. This process may sometimes result in the use of fractions, which indicates that the reactants decompose into the products in a ratio that is not a simple whole number.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions in chemistry. It provides a quantitative relationship between the amounts of substances involved in a reaction, which is pivotal for predicting yields and guiding the synthesis of compounds.

In decomposition reactions, stoichiometry helps us understand how much of a substance is needed to form a specific amount of product. For instance, in the reaction \(2\mathrm{LiHCO}_{3}(s) \rightarrow \mathrm{Li}_{2}\mathrm{CO}_{3}(s)+\mathrm{H}_{2}\mathrm{O}(g)+\frac{2}{3}\mathrm{CO}_{2}(g)\), stoichiometry allows us to calculate the amount of lithium bicarbonate needed to produce a certain mass of lithium carbonate, water, and carbon dioxide, taking into account the balanced coefficients.

Decomposition Reaction Chemistry

Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. Typically involving heat, light, or electricity (among other factors), these reactions are fundamental to understanding chemical processes like digestion and cellular respiration.

An example of a decomposition reaction is the breakdown of benzene, \(\mathrm{C}_{6}\mathrm{H}_{6}(l) \rightarrow 6\mathrm{C}(s)+3\mathrm{H}_{2}(g)\), into carbon and hydrogen gas. Understanding the intricacies of how these compounds decompose can provide insights into more complex chemical behavior and aid in various applications across scientific disciplines.

Chemical Reaction Types

Chemistry recognizes various types of chemical reactions, including but not limited to synthesis, decomposition, single replacement, double replacement, and combustion. Each type of reaction has its characteristics and balanced chemical equations that provide details on the reactants and products involved.

Decomposition reactions, as one of these types, involve the breakdown of compounds, as seen in the exercise with \(\operatorname{PBr}_{3}(l) \rightarrow \mathrm{P}_{4}(s)+\mathrm{Br}_{2}(l)\). This reaction type contrasts with synthesis reactions where simpler substances combine to form a more complex compound, showcasing the diversity and complexity of chemical processes.