Question

Balance each of the following equations that describe decomposition reactions. a. \(\mathrm{NI}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+\mathrm{I}_{2}(s)\) b. \(\operatorname{BaCO}_{3}(s) \rightarrow \operatorname{BaO}(s)+\mathrm{CO}_{2}(g)\) c. \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightarrow \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\operatorname{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{SO}_{4}(s) \rightarrow \mathrm{CuSO}_{4}(s)+\mathrm{NH}_{3}(g)\) e. \(\mathrm{NaN}_{3}(s) \rightarrow \mathrm{Na}_{3} \mathrm{N}(s)+\mathrm{N}_{2}(g)\)

Short answer

The balanced decomposition equations are as follows: a. \(2\mathrm{NI}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+3\mathrm{I}_{2}(s)\) b. \(\operatorname{BaCO}_{3}(s) \rightarrow\operatorname{BaO}(s)+\mathrm{CO}_{2}(g)\) c. \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightarrow 6\mathrm{C}(s) + 6\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\operatorname{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{SO}_{4}(s) \rightarrow \mathrm{CuSO}_{4}(s) + 4\mathrm{NH}_{3}(g)\) e. \(3\mathrm{NaN}_{3}(s) \rightarrow 2\mathrm{Na}_{3}\mathrm{N}(s) + \mathrm{N}_{2}(g)\)

Step-by-step solution

Balance Equation a: \(\mathrm{NI}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+\mathrm{I}_{2}(s)\)

We will count the number of atoms for each element on both sides of the equation. Reactant side: 1 N atom and 3 I atoms Product side: 2 N atoms and 2 I atoms Since there are 2 N atoms in the product but only 1 N atom in the reactant, we need to place a coefficient of 2 in front of \(\mathrm{NI}_{3}\). New count of Reactant side: 2 N atoms and 6 I atoms Now, since there are now 6 I atoms on the reactant side and only 2 I atoms on the product side, we need to place a coefficient of 3 in front of \(\mathrm{I}_{2}\). New count of Product side: 2 N atoms and 6 I atoms Balanced equation: \(2\mathrm{NI}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+3\mathrm{I}_{2}(s)\)

Balance Equation b: \(\operatorname{BaCO}_{3}(s) \rightarrow\operatorname{BaO}(s)+\mathrm{CO}_{2}(g)\)

Count the number of atoms for each element on both sides of the equation. Reactant side: 1 Ba atom, 1 C atom and 3 O atoms Product side: 1 Ba atom, 1 C atom and 2 O atoms There is one extra O atom on the reactant side; however, this equation is already balanced. Balanced equation: \(\operatorname{BaCO}_{3}(s) \rightarrow\operatorname{BaO}(s)+\mathrm{CO}_{2}(g)\)

Balance Equation c: \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightarrow\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g)\)

Count the number of atoms for each element on both sides of the equation. Reactant side: 6 C atoms, 12 H atoms and 6 O atoms Product side: 1 C atom, 2 H atoms and 1 O atom We need to place a coefficient of 6 in front of \(\mathrm{C}(s)\) and a coefficient of 6 in front of \(\mathrm{H}_{2} \mathrm{O}(g)\) to balance the atoms. Balanced equation: \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightarrow 6\mathrm{C}(s) + 6\mathrm{H}_{2} \mathrm{O}(g)\)

Balance Equation d: \(\operatorname{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{SO}_{4}(s) \rightarrow \mathrm{CuSO}_{4}(s)+\mathrm{NH}_{3}(g)\)

Count the number of atoms for each element on both sides of the equation. Reactant side: 1 Cu atom, 4 N atoms, 12 H atoms and 1 S atom and 4 O atoms Product side: 1 Cu atom, 1 N atom, 3 H atoms, 1 S atom and 4 O atoms We need to place a coefficient of 4 in front of \(\mathrm{NH}_{3}(g)\) to balance the atoms. Balanced equation: \(\operatorname{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{SO}_{4}(s) \rightarrow \mathrm{CuSO}_{4}(s) + 4\mathrm{NH}_{3}(g)\)

Balance Equation e: \(\mathrm{NaN}_{3}(s) \rightarrow \mathrm{Na}_{3}\mathrm{N}(s)+\mathrm{N}_{2}(g)\)

Count the number of atoms for each element on both sides of the equation. Reactant side: 1 Na atom, 3 N atoms Product side: 3 Na atoms and 4 N atoms We need to place a coefficient of 3 in front of \(\mathrm{NaN}_3\) and a coefficient of 2 in front of \(\mathrm{Na}_{3}\mathrm{N}(s)\) to balance the atoms. Balanced equation: \(3\mathrm{NaN}_{3}(s) \rightarrow 2\mathrm{Na}_{3}\mathrm{N}(s) + \mathrm{N}_{2}(g)\)

Key concepts

Decomposition Reactions

Decomposition reactions are a fundamental type of chemical reaction where a single compound breaks down into two or more simpler products. This type of reaction is characterized by being a kind of "reverse synthesis," where energy is usually needed for the compound to split into smaller products. This often happens when heat, light, or electricity is applied.

Consider a compound like \( ext{NaN}_3 (s)\), an explosive compound used in airbags, which decomposes into sodium and nitrogen gas: \(3 \text{NaN}_3(s) \rightarrow 2 \text{Na}_3\text{N}(s) + \text{N}_2(g)\). The decomposition of this compound releases nitrogen gas rapidly, providing the force needed to inflate an airbag.

• This process highlights how decomposition reactions can feature one reactant forming multiple products.
• Most decomposition reactions require an input of energy to occur, known as endothermic reactions.

Decomposition reactions are significant in fields like materials science and safety engineering, as compounds sometimes need to be broken down efficiently. Studying these reactions aids in understanding how substances can be transformed or utilized under specific conditions.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, required to accurately describe the ratios in which reactants combine to form products. An unbalanced equation might violate the law of conservation of mass, as atoms cannot be created or destroyed in a chemical reaction.

• The goal is to ensure the same number of each type of atom appears on both sides of the equation.
• Coefficients, which are numbers placed before compounds, are used to balance the equation.

To illustrate, consider the decomposition reaction of \(\text{C}_6\text{H}_{12}\text{O}_6(s)\). Initially, the equation may not balance as there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms in the reactant and differing numbers in the products. By adjusting the coefficients to balance the atoms,\(\text{C}_6\text{H}_{12}\text{O}_6(s) \rightarrow 6\text{C}(s) + 6\text{H}_2\text{O}(g)\), the equation aligns with the conservation of mass.

Balancing is crucial for stoichiometry and allows chemists to calculate exact amounts of reactants and products required.

Understanding Stoichiometry

Stoichiometry is the study of the quantitative relationships between the reactants and products in a chemical reaction. It fundamentally relies on balanced chemical equations, making balance crucial.

• It allows chemists to predict the amounts of products formed and the amounts of reactants required, based on the ratios derived from balanced equations.
• Understanding the stoichiometry in decomposition reactions helps in determining the efficiency and yield of the reaction.

For instance, in the breakdown of \(2\text{NI}_3(s) \rightarrow \text{N}_2(g) + 3\text{I}_2(s)\), stoichiometry tells us that 2 moles of nitrogen triiodide decompose to produce 1 mole of nitrogen gas and 3 moles of iodine.

By effectively using stoichiometry, chemical engineers design processes and derive the right quantities needed for reactions. This concept is essential not only in labs but also in industrial applications, from manufacturing pharmaceuticals to synthesizing materials.