Problem 14 - Chemistry and Chemical Reactivity

Question

Complete and balance the equations for the following reactions. (a) \(\mathrm{K}(\mathrm{s})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow\) (b) \(\mathrm{Ba}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow\) (c) \(\mathrm{Al}(\mathrm{s})+\mathrm{S}_{8}(\mathrm{s}) \longrightarrow\) (d) \(\mathrm{Si}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow\)

Step-by-Step Solution

Answer

(a) 2K + I₂ → 2KI; (b) 2Ba + O₂ → 2BaO; (c) 16Al + 3S₈ → 8Al₂S₃; (d) Si + 2Cl₂ → SiCl₄.

Step 1: Balance Reaction (a)

For the reaction between potassium (K) and iodine (I₂), potassium iodide (KI) is formed. The unbalanced equation is: \[ \mathrm{K} + \mathrm{I}_{2} \rightarrow \mathrm{KI} \]To balance it, note we have 2 iodine atoms on the reactant side, so we need 2 KI molecules on the product side. Balanced equation: \[ 2\mathrm{K} + \mathrm{I}_{2} \rightarrow 2\mathrm{KI} \]

Step 2: Balance Reaction (b)

For the reaction between barium (Ba) and oxygen (O₂), barium oxide (BaO) is formed. The unbalanced equation is: \[ \mathrm{Ba} + \mathrm{O}_{2} \rightarrow \mathrm{BaO} \]Since there are 2 oxygen atoms in \(\mathrm{O}_2\), we need to use 2 molecules of \(\mathrm{BaO}\) on the product side. Balanced equation:\[ 2\mathrm{Ba} + \mathrm{O}_{2} \rightarrow 2\mathrm{BaO} \]

Step 3: Balance Reaction (c)

For the reaction between aluminum (Al) and sulfur (S₈), aluminum sulfide (Al₂S₃) is formed. The unbalanced equation is: \[ \mathrm{Al} + \mathrm{S}_{8} \rightarrow \mathrm{Al}_2\mathrm{S}_3 \]Balance the sulfur by multiplying \(\mathrm{Al}_2\mathrm{S}_3\) by 4 to get \(12\) sulfur atoms on both sides. Then, balance aluminum by multiplying \(\mathrm{Al}\) by 8.Balanced equation:\[ 16\mathrm{Al} + 3\mathrm{S}_{8} \rightarrow 8\mathrm{Al}_2\mathrm{S}_3 \]

Step 4: Balance Reaction (d)

For the reaction between silicon (Si) and chlorine (Cl₂), silicon tetrachloride (SiCl₄) is formed. The unbalanced equation is: \[ \mathrm{Si} + \mathrm{Cl}_{2} \rightarrow \mathrm{SiCl}_4 \]Since \(\mathrm{SiCl}_4\) requires 4 chlorine atoms, use 2 \(\mathrm{Cl}_2\) molecules for every \(\mathrm{SiCl}_4\). Balanced equation:\[ \mathrm{Si} + 2\mathrm{Cl}_{2} \rightarrow \mathrm{SiCl}_4 \]

Key Concepts

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the reactants, which are substances consumed in the reaction, and the products, which are substances formed in the reaction. The left side of the equation shows the reactants, while the right side shows the products. A chemical equation is balanced when the number of atoms of each element is equal on both sides of the equation. This balance reflects the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction. In the given exercise, we had to complete and balance several chemical reactions such as the formation of potassium iodide from potassium and iodine, barium oxide from barium and oxygen, aluminum sulfide from aluminum and sulfur, and silicon tetrachloride from silicon and chlorine. Each equation must be balanced to ensure the Law of Conservation of Mass is maintained.

Stoichiometry

Stoichiometry is a core concept in chemistry that involves the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict how much of each substance is needed or produced in a reaction.To balance a chemical equation, like in our exercise, stoichiometric coefficients are used. These are the numbers placed before compounds in a chemical equation to ensure the same number of atoms for each element on both sides of the equation.For instance, in balancing the reaction between aluminum and sulfur, we determined that the stoichiometric coefficients were 16 for aluminum and 3 for sulfur in \[ 16\mathrm{Al} + 3\mathrm{S}_{8} \rightarrow 8\mathrm{Al}_2\mathrm{S}_3 \].This careful balance not only keeps the equation consistent with the Law of Conservation of Mass, but it also provides essential information for calculating amounts of reactants required or products formed.

Inorganic Chemistry

Inorganic chemistry is the study of inorganic compounds, typically those that do not contain carbon-hydrogen bonds. The reactions in our exercise are all examples of inorganic chemical reactions, where elements combine to form compounds. For instance, when potassium and iodine react, they form potassium iodide, a simple ionic compound. Similarly, the reaction of silicon and chlorine to form silicon tetrachloride showcases the typical formation of halides in inorganic reactions. Inorganic reactions often involve metals and nonmetals and can feature a wide variety of compounds, from ionic salts to covalently bonded molecules. Understanding these reactions is crucial, as they happen frequently in industrial processes and natural phenomena. In our exercise, each balanced equation reflects these typical inorganic processes, showcasing the diverse ways elements can combine and the importance of proper balancing in representing these changes accurately.