Question

Balance each of the following decomposition reactions: (a) \(\mathrm{NaClO}_{3}(s) \rightarrow \mathrm{NaCl}(s)+\mathrm{O}_{2}(g)\) (b) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s) \rightarrow \mathrm{Ca}\left(\mathrm{NO}_{2}\right)_{2}(s)+\mathrm{O}_{2}(g)\)

Short answer

(a) \(2\mathrm{NaClO}_{3} \rightarrow 2\mathrm{NaCl} + 3\mathrm{O}_{2}\); (b) \(2\mathrm{Ca}(\mathrm{NO}_{3})_{2} \rightarrow 2\mathrm{Ca}(\mathrm{NO}_{2})_{2} + 2\mathrm{O}_{2}\).

Step-by-step solution

Identify the Initial Equation

For reaction (a), the equation is \(\mathrm{NaClO}_{3}(s) \rightarrow \mathrm{NaCl}(s)+\mathrm{O}_{2}(g)\). The goal is to balance both sides to have the same number of each type of atom.

Balance Sodium, Chlorine, and Oxygen in (a)

Start by balancing one element at a time. Na and Cl atoms are already balanced with one each on both sides. For oxygen, there are three in \(\mathrm{NaClO}_{3}\) and we need to make sure the total on the right matches. This will require using the smallest whole number multiples to balance O atoms. Adjust the coefficients to get: \(2\mathrm{NaClO}_{3}(s) \rightarrow 2\mathrm{NaCl}(s) + 3\mathrm{O}_{2}(g)\). There are now 6 oxygen atoms on each side.

Identify the Initial Equation for Reaction (b)

For reaction (b), the equation is \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s) \rightarrow \mathrm{Ca}\left(\mathrm{NO}_{2}\right)_{2}(s) + \mathrm{O}_{2}(g)\). We will balance elements sequentially.

Balance Calcium, Nitrogen, and Oxygen in (b)

Calcium and nitrogen are already balanced since there is one \(\mathrm{Ca}\) and two \(\mathrm{NO}\) fixtures on both sides. Focus on oxygen. Initially, there are 6 oxygen atoms in \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) on the left. We need to add enough \(\mathrm{O}_{2}\) molecules to balance the equation. Each \(\mathrm{O}_{2}\) molecule adds 2 oxygen atoms to the right-hand side. This can be achieved using the equation: \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s) \rightarrow \mathrm{Ca}\left(\mathrm{NO}_{2}\right)_{2}(s) + \mathrm{O}_{2}(g)\), where \(\frac{1}{2}\) \(\mathrm{O}_{2}\) was initially incorrectly stated.

Include Correct Coefficients to Finalize (b)

In actuality, to simplify, instead of using fractions, multiply the entire equation by 2: \(2\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s) \rightarrow 2\mathrm{Ca}\left(\mathrm{NO}_{2}\right)_{2}(s) + 2\mathrm{O}_{2}(g)\). Now, check all elements to ensure they are balanced across the reaction.

Key concepts

Understanding Decomposition Reactions

Decomposition reactions are a type of chemical reaction in which one compound breaks down into two or more simpler substances. A classic example is the breakdown of water (H_2O) into hydrogen and oxygen gases. In the given exercise, we observed the decomposition of salts such as sodium chlorate (NaClO_3) and calcium nitrate (Ca(NO_3)_2).

• In reaction (a), sodium chlorate decomposes into sodium chloride and oxygen gas.
• For reaction (b), calcium nitrate breaks down into calcium nitrite and oxygen gas.

Decomposition reactions are often identified by a single reactant turning into multiple products. It usually requires some form of energy input, such as heat or light. Understanding these reactions helps us grasp how complex compounds can derive into simpler elements or molecules.

Balancing Chemical Equations in Decomposition Reactions

Balancing chemical equations is essential to comply with the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. This means the number of atoms for each element must be the same on both sides of the equation. In the original exercise, we were presented with decomposition reactions that needed balancing. Let's explore how this is achieved:

• Sodium and Chlorine Balance: In reaction (a), both sodium and chlorine atoms were already balanced with 1 atom on each side, so our focus was directed at oxygen.
• Oxygen Balance: Initially, we had 3 oxygen atoms in NaClO_3. By using a coefficient of 2 for NaClO_3 and 3 for O_2, six oxygen atoms were equalized on both sides.
• Calcium and Nitrogen Balance: In reaction (b), the calcium and nitrogen atoms were balanced, leaving oxygen as the primary atom to adjust. The correct coefficients overall became 2 for both initial and all resulting products to ensure a balanced equation across the board.

Balancing is often a trial and error process involving adjusting coefficients of reactants and products until all atoms are balanced. It's vital since it aids in correct stoichiometry calculations.

Grasping Stoichiometry Basics

Stoichiometry is the calculation of reactants and products in chemical reactions. This concept can be quite abstract unless everything is balanced properly, hence the importance of our previous step. Once an equation is balanced, stoichiometry facilitates:

• Predicting Product Quantities: By determining the relative amounts of reactants, we can predict the quantities of products formed.
• Scaling Reactions: Stoichiometry allows for the scaling of a chemical reaction to any desired quantity, ensuring all reactants and products remain proportionate.
• Unknown Quantification: Should a particular reactant's amount be unknown, stoichiometry will enable computation based on other known quantities.

Stoichiometry highlights the quantitative aspects of chemical reactions and provides the foundation for understanding reaction yields and limiting reactants, which are pivotal in many scientific and industrial applications.