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) \(\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow\)

Short answer

(a) \(\mathrm{CaO}(s) + 2\mathrm{H_2O}(l) \longrightarrow \mathrm{Ca(OH)_2}(aq)\) (b) \(\mathrm{Al_2O_3}(s) + 6\mathrm{H^+}(aq) \longrightarrow 2\mathrm{Al^{3+}}(aq) + 3\mathrm{H_2O}(l)\) (c) \(\mathrm{Na_2O_2}(s) + 2\mathrm{H_2O}(l) \longrightarrow 2\mathrm{NaOH}(aq) + \mathrm{H_2O_2}(aq)\) (d) \(\mathrm{N_2O_3}(g) + \mathrm{H_2O}(l) \longrightarrow 2\mathrm{HNO_2}(aq)\) (e) \(\mathrm{KO_2}(s) + 2\mathrm{H_2O}(l) \longrightarrow 2\mathrm{KOH}(aq) + \mathrm{H_2O_2}(aq) + \mathrm{O_2}(g)\) (f) \(2\mathrm{NO}(g) + \mathrm{O_3}(g) \longrightarrow \mathrm{N_2O_5}(g)\)

Step-by-step solution

(a) Balancing the CaO and H2O equation

To balance this equation, first identify the number of atoms of each element on both sides of the equation. Ca and O are balanced, but there's 1 H atom on the left and 2 H atoms on the right. To balance the H atoms, add a coefficient of 2 in front of the H2O on the left side. The balanced equation is: \(\mathrm{CaO}(s) + 2\mathrm{H_2O}(l) \longrightarrow \mathrm{Ca(OH)_2}(aq)\)

(b) Balancing the Al2O3 and H+ equation

Balance the Al and O atoms. Add a coefficient of 6 in front of the H+ on the right side to balance the O atoms. Now there are 3 Al atoms on the left and 6 O atoms on both sides. Thus, the balanced equation is: \(\mathrm{Al_2O_3}(s) + 6\mathrm{H^+}(aq) \longrightarrow 2\mathrm{Al^{3+}}(aq) + 3\mathrm{H_2O}(l)\)

(c) Balancing the Na2O2 and H2O equation

Balance the Na, O, and H atoms. Add a coefficient of 2 in front of the H2O on the right side to balance the O atoms. Now there are 2 Na atoms on both sides and 2 H atoms on both sides. The balanced equation is: \(\mathrm{Na_2O_2}(s) + 2\mathrm{H_2O}(l) \longrightarrow 2\mathrm{NaOH}(aq) + \mathrm{H_2O_2}(aq)\)

(d) Balancing the N2O3 and H2O equation

Balance the N, O, and H atoms. Add a coefficient of 2 in front of the H2O on the right side to balance the N atoms. Now there are 2 N atoms and 5 O atoms on both sides. Thus, the balanced equation is: \(\mathrm{N_2O_3}(g) + \mathrm{H_2O}(l) \longrightarrow 2\mathrm{HNO_2}(aq)\)

(e) Balancing the KO2 and H2O equation

Balance the K, O, and H atoms. Add a coefficient of 2 in front of the H2O on the right side to balance the O atoms. Now there are 2 K atoms on both sides and 2 H atoms on both sides. The balanced equation is: \(\mathrm{KO_2}(s) + 2\mathrm{H_2O}(l) \longrightarrow 2\mathrm{KOH}(aq) + \mathrm{H_2O_2}(aq) + \mathrm{O_2}(g)\)

(f) Balancing the NO and O3 equation

Balance the N and O atoms. Add a coefficient of 2 in front of the NO on the left side to balance the N atoms. Now there are 2 N atoms and 5 O atoms on both sides. Thus, the balanced equation is: \(2\mathrm{NO}(g) + \mathrm{O_3}(g) \longrightarrow \mathrm{N_2O_5}(g)\)

Key concepts

Chemical Equations

In chemistry, chemical equations are a symbolic representation of chemical reactions. They show the reactants on the left and the products on the right, joined by an arrow indicating the direction of the reaction. Each chemical equation must be balanced to adhere to the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction. This means the number and types of atoms on each side must be equal.

Balancing these equations requires adjusting the coefficients (the numbers in front of compounds) to ensure that the same amount of each element is present on both sides. For example, in the balanced equation for the reaction of calcium oxide with water:

• The unbalanced equation: \(\mathrm{CaO}(s) + \mathrm{H_2O}(l) \longrightarrow \)
• The balanced equation: \(\mathrm{CaO}(s) + 2\mathrm{H_2O}(l) \longrightarrow \mathrm{Ca(OH)_2}(aq)\)

The number of each type of atom (Ca, O, and H) is the same on both sides, ensuring the equation respects the conservation of mass.

Stoichiometry

Stoichiometry refers to the calculation of reactants and products in chemical reactions. It involves using balanced chemical equations to predict how much of a substance will react or form. Key to stoichiometry is the mole concept, allowing chemists to convert between moles of different substances.

For instance, in the reaction \(\mathrm{Na_2O_2}(s) + 2\mathrm{H_2O}(l) \longrightarrow 2\mathrm{NaOH}(aq) + \mathrm{H_2O_2}(aq)\), stoichiometry tells us how moles of water will produce a certain amount of sodium hydroxide and hydrogen peroxide. By understanding the stoichiometric ratios (as given by the coefficients in a balanced equation), calculations such as the amount of reactant needed or product formed can be precisely determined. Stoichiometry allows us to scale reactions to be feasible for laboratory or industrial processes.

Inorganic Chemistry

Inorganic chemistry primarily deals with compounds not categorized as organic, typically including the vast range of elements besides carbon. The realm of inorganic chemistry encompasses reactions involving metals, salts, and other non-carbon-based substances.

Each of the given reactions in the exercise, such as the reaction between aluminum oxide \(\text{Al}_2\text{O}_3\) and hydrogen ions \(\text{H}^+\), exemplifies this field. Inorganic reactions often involve the transfer of ions and electrons and include the formation of salts, such as when aluminum ions form in solution. This field is crucial for creating materials like metals, catalysts, and ceramics. Understanding these reactions provides valuable insights into material science and industrial chemistry.

Acid-Base Reactions

Acid-base reactions are a subset of chemical reactions involving an acid (proton donor) and a base (proton acceptor). These reactions are fundamental in both industrial applications and natural processes, including the digestion of food and the functioning of batteries.

Consider one exercise reaction: \(\mathrm{Al_2O_3}(s) + 6\mathrm{H^+}(aq) \longrightarrow 2\mathrm{Al^{3+}}(aq) + 3\mathrm{H_2O}(l)\). Here, hydrogen ions act as acid, reacting with aluminum oxide, a base, to produce aluminum ions and water.

• The acid in the reaction donates protons to the base.
• The base accepts the protons, often resulting in the formation of water.

Grasping acid-base reactions is vital for understanding processes in both the laboratory and everyday life, such as in maintaining pH balance in chemical processes.