Problem 81 - Chemistry The Central Science

Question

Write a balanced equation for the reaction of each of the following compounds with water: (a) \(\mathrm{SO}_{2}(g),(\mathbf{b}) \mathrm{Cl}_{2} \mathrm{O}_{7}(g)\) (c) \(\mathrm{Na}_{2} \mathrm{O}_{2}(s),\) (d) \(\mathrm{BaC}_{2}(s),\) (e) \(\mathrm{RbO}_{2}(s)\) (f) \(\mathrm{Mg}_{3} \mathrm{N}_{2}(s)\) , (g) \(\mathrm{NaH}(s) .\)

Step-by-Step Solution

Answer

Here are the balanced equations for the reactions of the given compounds with water: (a) \(\mathrm{SO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{H}_{2}\mathrm{SO}_{3}(aq)\) (b) \(\mathrm{Cl}_{2}\mathrm{O}_{7}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 2\mathrm{HClO}_{4}(aq)\) (c) \(\mathrm{Na}_{2}\mathrm{O}_{2}(s) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 2\mathrm{NaOH}(aq) + \mathrm{O}_{2}(g)\) (d) \(\mathrm{BaC}_{2}(s) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{Ba(OH)}_{2}(s) + \mathrm{C}_{2}\mathrm{H}_{2}(g)\) (e) \(\mathrm{RbO}_{2}(s) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 2\mathrm{RbOH}(aq) + \mathrm{O}_{2}(g)\) (f) \(\mathrm{Mg}_{3}\mathrm{N}_{2}(s) + 6\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 3\mathrm{Mg(OH)}_{2}(s) + 2\mathrm{NH}_{3}(g)\) (g) \(\mathrm{NaH}(s) + \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{NaOH}(aq) + \mathrm{H}_{2}(g)\)

Step 1: (a) Reaction of \(\mathrm{SO}_{2}(g)\) with water

When sulfur dioxide (\(\mathrm{SO}_{2}\)) reacts with water (\(\mathrm{H}_2\mathrm{O}\)), it forms sulfurous acid (\(\mathrm{H}_2\mathrm{SO}_{3}\)). The balanced chemical equation for this reaction is: \[\mathrm{SO}_{2}(g) +\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{H}_{2}\mathrm{SO}_{3}(aq)\]

Step 2: (b) Reaction of \(\mathrm{Cl}_{2}\mathrm{O}_{7}(g)\) with water

When dichlorine heptoxide (\(\mathrm{Cl}_{2}\mathrm{O}_{7}\)) reacts with water (\(\mathrm{H}_2\mathrm{O}\)), it forms perchloric acid (\(\mathrm{HClO}_{4}\)). The balanced chemical equation for this reaction is: \[\mathrm{Cl}_{2}\mathrm{O}_{7}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 2\mathrm{HClO}_{4}(aq)\]

Step 3: (c) Reaction of \(\mathrm{Na}_{2}\mathrm{O}_{2}(s)\) with water

When sodium peroxide (\(\mathrm{Na}_{2}\mathrm{O}_{2}\)) reacts with water (\(\mathrm{H}_2\mathrm{O}\)), it forms sodium hydroxide (\(\mathrm{NaOH}\)) and oxygen gas (\(\mathrm{O}_{2}\)). The balanced chemical equation for this reaction is: \[\mathrm{Na}_{2}\mathrm{O}_{2}(s) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 2\mathrm{NaOH}(aq) + \mathrm{O}_{2}(g)\]

Step 4: (d) Reaction of \(\mathrm{BaC}_{2}(s)\) with water

When barium carbide (\(\mathrm{BaC}_{2}\)) reacts with water (\(\mathrm{H}_2\mathrm{O}\)), it forms barium hydroxide (\(\mathrm{Ba(OH)}_{2}\)) and acetylene (\(\mathrm{C}_{2}\mathrm{H}_{2}\)). The balanced chemical equation for this reaction is: \[\mathrm{BaC}_{2}(s) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{Ba(OH)}_{2}(s) + \mathrm{C}_{2}\mathrm{H}_{2}(g)\]

Step 5: (e) Reaction of \(\mathrm{RbO}_{2}(s)\) with water

When rubidium superoxide (\(\mathrm{RbO}_{2}\)) reacts with water (\(\mathrm{H}_2\mathrm{O}\)), it forms rubidium hydroxide (\(\mathrm{RbOH}\)) and oxygen gas (\(\mathrm{O}_{2}\)). The balanced chemical equation for this reaction is: \[\mathrm{RbO}_{2}(s) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 2\mathrm{RbOH}(aq) + \mathrm{O}_{2}(g)\]

Step 6: (f) Reaction of \(\mathrm{Mg}_{3}\mathrm{N}_{2}(s)\) with water

When magnesium nitride (\(\mathrm{Mg}_{3}\mathrm{N}_{2}\)) reacts with water (\(\mathrm{H}_2\mathrm{O}\)), it forms magnesium hydroxide (\(\mathrm{Mg(OH)}_{2}\)) and ammonia (\(\mathrm{NH}_{3}\)). The balanced chemical equation for this reaction is: \[\mathrm{Mg}_{3}\mathrm{N}_{2}(s) + 6\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 3\mathrm{Mg(OH)}_{2}(s) + 2\mathrm{NH}_{3}(g)\]

Step 7: (g) Reaction of \(\mathrm{NaH}(s)\) with water

When sodium hydride (\(\mathrm{NaH}\)) reacts with water (\(\mathrm{H}_2\mathrm{O}\)), it forms sodium hydroxide (\(\mathrm{NaOH}\)) and hydrogen gas (\(\mathrm{H}_{2}\)). The balanced chemical equation for this reaction is: \[\mathrm{NaH}(s) + \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{NaOH}(aq) + \mathrm{H}_{2}(g)\]

Key Concepts

balanced chemical equations

A balanced chemical equation is crucial for representing a chemical reaction accurately. It shows the exact number of atoms of each element involved in the reaction, ensuring that mass is conserved. This is known as the Law of Conservation of Mass. For instance, when balancing the reaction of \( \mathrm{SO}_2(g) \) with water, the equation becomes: \( \mathrm{SO}_2(g) + \mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{H}_2\mathrm{SO}_3(aq) \).

• Each side of the equation contains equal numbers of each type of atom.
• Balancing involves adding or changing coefficients, which are the numbers before molecules.

If you find that you are adding or changing subscripts within a chemical formula, you are altering the compounds, which is not correct. Remember, balancing equations ensures that the chemical process is accurately depicted, respecting both the quantitative and qualitative aspects of the substances involved.

reactions with water

Reactions with water, also called hydrolysis reactions, are a common type of reaction in inorganic chemistry. Water can interact with other compounds to form acids, bases, or evolve gases. For example:- When dichlorine heptoxide, \( \mathrm{Cl}_2\mathrm{O}_7 \), reacts with water, it forms a strong acid: \( \mathrm{Cl}_2\mathrm{O}_7(g) + 2\mathrm{H}_2\mathrm{O}(l) \rightarrow 2\mathrm{HClO}_4(aq) \).In this reaction, water's role is to facilitate the conversion of an oxide, \( \mathrm{Cl}_2\mathrm{O}_7 \), into perchloric acid.- Likewise, water helps transform sodium peroxide into a base and oxygen gas: \( \mathrm{Na}_2\mathrm{O}_2(s) + 2\mathrm{H}_2\mathrm{O}(l) \rightarrow 2\mathrm{NaOH}(aq) + \mathrm{O}_2(g) \).Understanding these reactions requires recognizing water’s versatility in forming different compounds through chemical reactions.

compound reactions

Compound reactions involve the transformation of compounds into different substances. These transformations may involve synthesis, decomposition, single replacement, or double replacement reactions. In the exercise examples, several synthesis reactions occur, such as:- Barium carbide reacting with water to produce barium hydroxide and acetylene gas, a synthesis of new compounds: \( \mathrm{BaC}_2(s) + 2\mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{Ba(OH)}_2(s) + \mathrm{C}_2\mathrm{H}_2(g) \).In compound reactions, understanding the products formed helps predict the type of reaction. Another example is the reaction of magnesium nitride with water, forming magnesium hydroxide and ammonia: \( \mathrm{Mg}_3\mathrm{N}_2(s) + 6\mathrm{H}_2\mathrm{O}(l) \rightarrow 3\mathrm{Mg(OH)}_2(s) + 2\mathrm{NH}_3(g) \).Knowing these specifics can assist in classifying reactions and understanding their broader context in chemistry.

inorganic chemistry

Inorganic chemistry focuses on the behavior and synthesis of inorganic compounds, typically those that do not contain carbon-hydrogen bonds. These can include metals, minerals, salts, and more. It forms the foundation for understanding various reactions, such as those involving ionic compounds or transition metals. For example:- Sodium hydride, \( \mathrm{NaH} \), reacts with water to produce a base and hydrogen gas: \( \mathrm{NaH}(s) + \mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{NaOH}(aq) + \mathrm{H}_2(g) \).This is characteristic of many reactions in inorganic chemistry, where active metals or their compounds react with water to release hydrogen gas. Understanding these elemental interactions in inorganic chemistry is essential for exploring more complex chemical reactions and materials synthesis.