Problem 38 - Chemistry: The Central Science in SI Units, Global Edition

Question

Complete and balance the following equations: (a) \(\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow\) (c) \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (d) \(\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (e) \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (f) \(\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow\)

Step-by-Step Solution

Answer

(a) \(\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca(OH)}_2\) (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+6\mathrm{H}^{+}(a q) \longrightarrow 2\mathrm{Al}^{3+}(aq) + 3\mathrm{H_2O}\) (c) \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{NaOH}(aq) +\frac{1}{2}\mathrm{O}_{2}(g)\) (d) \(\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{HNO_2}\) (e) \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{KOH}(aq) + \mathrm{H_2O_2}(l) + \frac{1}{2}\mathrm{O}_{2}(g)\) (f) \(\mathrm{NO}(g)+\mathrm{O_{3}}(g) \longrightarrow \mathrm{NO}_2(g) + \mathrm{O}_2(g)\)

Step 1: In this reaction, \(\mathrm{CaO}\) (calcium oxide) reacts with \(\mathrm{H}_{2} \mathrm{O}\) (water). Since \(\mathrm{CaO}\) is a metal oxide, it will react with water to produce a metal hydroxide, which in this case is \(\mathrm{Ca(OH)}_2\) (calcium hydroxide). (a) \(\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca(OH)}_2\) #Step 2: Balance the equation#

Now, we count the number of atoms of each element on both sides of the equation. Here, we have 1 \(\mathrm{Ca}\), 1 \(\mathrm{O}\), and 2 \(\mathrm{H}\) on both sides of the equation. This equation is already balanced. \( \boxed{\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca(OH)}_2} \) (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow\) #Step 1: Identify the product#

Step 2: In this reaction, \(\mathrm{Al}_2 \mathrm{O}_{3}\) (aluminum oxide) reacts with \(\mathrm{H}^{+}\) ions in an aqueous solution. Here, we will form an aluminum salt, which is \(\mathrm{Al}^{3+}\) and \(\mathrm{H_2O}\). (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Al}^{3+}(aq) + \mathrm{H_2O}\) #Step 2: Balance the equation#

We require 6 \(\mathrm{H}^{+}\) ions and 3 \(\mathrm{H_2O}\) molecules to balance the number of atoms in the reactants and products. Therefore, the balanced equation is: \( \boxed{\mathrm{Al}_{2} \mathrm{O}_{3}(s)+6\mathrm{H}^{+}(a q) \longrightarrow 2\mathrm{Al}^{3+}(aq) + 3\mathrm{H_2O}} \) (c) \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) #Step 1: Identify the product#

Step 3: In this reaction, \(\mathrm{Na}_2 \mathrm{O}_2\) (sodium peroxide) reacts with \(\mathrm{H}_2 \mathrm{O}\) (water). Sodium peroxide will decompose in water, forming sodium hydroxide and oxygen gas. (c) \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{NaOH}(aq) +\frac{1}{2}\mathrm{O}_{2}(g)\) #Step 2: Balance the equation#

In this case, the equation is already balanced since we have 2 \(\mathrm{Na}\), 4 \(\mathrm{O}\), and 2 \(\mathrm{H}\) on both sides. \( \boxed{\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{NaOH}(aq) +\frac{1}{2}\mathrm{O}_{2}(g)} \) (d) \(\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) #Step 1: Identify the product#

Step 4: In this reaction, \(\mathrm{N}_{2} \mathrm{O}_{3}\) (dinitrogen trioxide) reacts with \(\mathrm{H}_2 \mathrm{O}\) (water). Dinitrogen trioxide reacts with water to produce nitrous acid (\(\mathrm{HNO}_2\)). (d) \(\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{HNO_2}\) #Step 2: Balance the equation#

We now have 2 \(\mathrm{N}\), 6 \(\mathrm{O}\), and 4 \(\mathrm{H}\) atoms on both sides of the equation, so it is balanced. \( \boxed{\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{HNO_2}} \) (e) \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) #Step 1: Identify the product#

Step 5: In this reaction, \(\mathrm{KO}_{2}\) (potassium superoxide) reacts with \(\mathrm{H}_{2} \mathrm{O}\) (water). Potassium superoxide will decompose in water, forming potassium hydroxide, hydrogen peroxide, and oxygen gas. (e) \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{KOH}(aq) + \mathrm{H_2O_2}(l) + \frac{1}{2}\mathrm{O}_{2}(g)\) #Step 2: Balance the equation#

In this case, the equation is already balanced since we have 1 \(\mathrm{K}\), 4 \(\mathrm{O}\), and 4 \(\mathrm{H}\) on both sides. \( \boxed{\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{KOH}(aq) + \mathrm{H_2O_2}(l) + \frac{1}{2}\mathrm{O}_{2}(g)} \) (f) \(\mathrm{NO}(g)+\mathrm{O_{3}}(g) \longrightarrow\) #Step 1: Identify the product#

Step 6: In this reaction, \(\mathrm{NO}\) (nitric oxide) reacts with \(\mathrm{O}_3\) (ozone). In this reaction, nitric oxide and ozone will produce nitrogen dioxide and molecular oxygen. (f) \(\mathrm{NO}(g)+\mathrm{O_{3}}(g) \longrightarrow \mathrm{NO}_2(g) + \mathrm{O}_2(g)\) #Step 2: Balance the equation#

In this case, the equation is already balanced since we have 1 \(\mathrm{N}\), 4 \(\mathrm{O}\) atoms on both sides. \( \boxed{\mathrm{NO}(g)+\mathrm{O_{3}}(g) \longrightarrow \mathrm{NO}_2(g) + \mathrm{O}_2(g)} \)

Key Concepts

Balancing Equations

Chemical equations need to be balanced to obey the Law of Conservation of Mass. This law states that matter cannot be created or destroyed in a chemical reaction, which means the number of atoms of each element must be the same on the reactant side as on the product side.
To achieve this, we adjust the coefficients (the numbers before molecules in a chemical equation) but never change the subscripts in the chemical formula.
This ensures that each atom in the reactants corresponds to an atom in the products, maintaining equilibrium.

• Start by balancing elements appearing in only one reactant and product.
• Next, balance elements present in more than one reactant or product.
• Finally, balance the oxygen and hydrogen atoms, often easier to balance at the end due to their presence in multiple compounds.

Balancing equations requires practice and can initially be challenging. It becomes much simpler with experience. Remember, a balanced equation reflects the true nature of the chemical reaction in reality.

Metal Oxides

Metal oxides are compounds consisting of metal atoms bonded with oxygen atoms. These are often formed when metals combine with oxygen in processes like rusting or burning.
Metal oxides generally react with water to form metal hydroxides. For example, calcium oxide (\(\mathrm{CaO}\)) reacts with water to produce calcium hydroxide (\(\mathrm{Ca(OH)}_2\)). This reaction is typical for many metal oxides:

• Some metal oxides act as basic oxides and react with acids in neutralization reactions, forming salts and water.
• Transition metal oxides might exhibit amphoteric behavior, meaning they can react with both acids and bases, like aluminum oxide (\(\mathrm{Al}_2\mathrm{O}_3\)).

Understanding the behavior of metal oxides is essential in predicting the outcomes of their reactions with other substances.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, involve the transfer of electrons from one substance to another. In these reactions, one reactant is oxidized (loses electrons), while another is reduced (gains electrons).
These processes are crucial in energy generation, such as in biological respiration and combustion. To identify redox reactions:

• Identify changes in oxidation states of elements in the reactants and products.
• Determine which element is being oxidized and which is being reduced— the element that increases in oxidation state is oxidized, and the one that decreases is reduced.

A classic redox reaction is between nitric oxide (\(\mathrm{NO}\)) and ozone (\(\mathrm{O}_3\)), where \(\mathrm{NO}\) gets oxidized to \(\mathrm{NO}_2\), and \(\mathrm{O}_3\) is reduced to \(\mathrm{O}_2\). This understanding is vital for comprehending many natural and industrial processes.

Acid-Base Reactions

Acid-base reactions are fundamental chemical processes involving the transfer of protons (\(\mathrm{H}^+\)) between reactants. In these reactions, an acid donates a proton, and a base accepts it. This exchange produces water and a salt.
For example, when aluminum oxide reacts with acidic protons (\(\mathrm{H}^+\)), it forms aluminum ions and water:

• These reactions play critical roles in daily life, from digestion in the human body to industrial manufacturing like soap production.
• Acid-base reactions are also integral in buffering solutions, which maintain constant pH in various environments, crucial for biochemical processes.

Grasping how acids and bases interact is essential for predicting the outcomes of these reactions in both laboratory and real-world scenarios.