Question

Complete and balance the following equations: (a) \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (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) \(\mathrm{NO}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow\)

Short answer

(a) \(CaO(s) + H_2O(l) \longrightarrow Ca(OH)_2\) (b) \(Al_2O_3(s) + 6H^+(aq) \longrightarrow 2Al^{3+}(aq) + 3H_2O(l)\) (c) \(Na_2O_2(s) + 2H_2O(l) \longrightarrow 2NaOH(aq) + O_2(g)\) (d) \(N_2O_3(g) + H_2O(l) \longrightarrow 2HNO_2(aq)\) (e) \(2KO_2(s) + 2H_2O(l) \longrightarrow 2KOH(aq) + H_2O_2(aq) + O_2(g)\) (f) \(NO(g) + O_3(g) \longrightarrow NO_2(g) + O_2(g)\)

Step-by-step solution

(a) Complete the reaction for CaO(s) + H2O(l)

The given equation is: \(CaO(s) + H_2O(l) \longrightarrow \) When calcium oxide (CaO) reacts with water (H2O), it forms calcium hydroxide (Ca(OH)2). So, the completed equation becomes: \(CaO(s) + H_2O(l) \longrightarrow Ca(OH)_2 \) Now, let's balance the equation by making sure that the number of atoms of each element on both sides of the equation is equal. \(CaO(s) + H_2O(l) \longrightarrow 1 Ca(OH)_2\) In this case, since the number of atoms of each element on both sides of the equation is already equal, the given equation is balanced.

(b) Complete the reaction for Al2O3(s) + H+(aq)

The given equation is: \(Al_2O_3(s) + H^+(aq) \longrightarrow \) When aluminum oxide (Al2O3) reacts with hydrogen ions (H+), it forms aluminum ions (Al3+) and water (H2O). So, the completed equation becomes: \(Al_2O_3(s) + H^+(aq) \longrightarrow Al^{3+}(aq) + H_2O(l) \) Now, let's balance the equation: \[Al_2O_3(s) + 6 H^+(aq) \longrightarrow 2 Al^{3+}(aq) + 3 H_2O(l)\] Balancing the equation, we get: \(Al_2O_3(s) + 6H^+(aq) \longrightarrow 2Al^{3+}(aq) + 3H_2O(l)\)

(c) Complete the reaction for Na2O2(s) + H2O(l)

The given equation is: \(Na_2O_2(s) + H_2O(l) \longrightarrow \) When sodium peroxide (Na2O2) reacts with water (H2O), it forms sodium hydroxide (NaOH) and oxygen (O2). So, the completed equation becomes: \(Na_2O_2(s) + H_2O(l) \longrightarrow NaOH(aq) + O_2(g) \) Now, let's balance the equation: \[Na_2O_2(s) + 2 H_2O(l) \longrightarrow 2 NaOH(aq) + O_2(g) \] Balancing the equation, we get: \(Na_2O_2(s) + 2H_2O(l) \longrightarrow 2NaOH(aq) + O_2(g)\)

(d) Complete the reaction for N2O3(g) + H2O(l)

The given equation is: \(N_2O_3(g) + H_2O(l) \longrightarrow \) When dinitrogen trioxide (N2O3) reacts with water (H2O), it forms nitrous acid (HNO2). So, the completed equation becomes: \(N_2O_3(g) + H_2O(l) \longrightarrow HNO_2(aq) \) Now, let's balance the equation: \[N_2O_3(g) + H_2O(l) \longrightarrow 2 HNO_2(aq)\] Balancing the equation, we get: \(N_2O_3(g) + H_2O(l) \longrightarrow 2HNO_2(aq)\)

(e) Complete the reaction for KO2(s) + H2O(l)

The given equation is: \(KO_2(s) + H_2O(l) \longrightarrow \) When potassium superoxide (KO2) reacts with water (H2O), it forms potassium hydroxide (KOH), hydrogen peroxide (H2O2), and oxygen gas (O2). So, the completed equation becomes: \(KO_2(s) + H_2O(l) \longrightarrow KOH(aq) + H_2O_2(aq) + O_2(g) \) Now, let's balance the equation: \[2 KO_2(s) + 2 H_2O(l) \longrightarrow 2 KOH(aq) + H_2O_2(aq) + O_2(g) \] Balancing the equation, we get: \(2KO_2(s) + 2H_2O(l) \longrightarrow 2KOH(aq) + H_2O_2(aq) + O_2(g)\)

(f) Complete the reaction for NO(g) + O3(g)

The given equation is: \(NO(g) + O_3(g) \longrightarrow \) When nitrogen monoxide (NO) reacts with ozone (O3), it forms nitrogen dioxide (NO2) and diatomic oxygen (O2). So, the completed equation becomes: \( NO(g) + O_3(g) \longrightarrow NO_2(g) + O_2(g) \) Now, let's balance the equation. In this case, since the number of atoms of each element on both sides of the equation is already equal, the given equation is balanced: \[NO(g) + O_3(g) \longrightarrow NO_2(g) + O_2(g) \]

Key concepts

Understanding Chemical Reactions

Chemical reactions are the processes by which substances, known as reactants, transform into new substances, called products. These transformations occur as a result of the breaking and forming of chemical bonds involving the movement of electrons. Understanding chemical reactions is a fundamental part of chemistry since they underpin many natural processes and technological applications.

The reactions presented in the exercise are examples of combination and decomposition reactions. In each, reactants combine or decompose to yield various products, which are signified through a chemical equation. The initial step in understanding these reactions is to write out the reactants and predict the products based on the chemical characteristics and reactivity of the involved compounds. For example, when calcium oxide reacts with water, it produces calcium hydroxide, a base commonly used in agriculture and construction.

Ensuring that each element is conserved in a reaction is central to the Law of Conservation of Mass. This law states that matter is neither created nor destroyed in a chemical reaction which is why it is important to balance chemical equations accurately. It is the ability to anticipate the outcome of a reaction and the necessity to balance the equation that aids in the deeper understanding of chemical processes.

The Role of Stoichiometry in Chemical Equations

Stoichiometry is the quantitative study of the relative amounts of reactants and products in chemical reactions. It is based on the conservation of mass and the concept of the mole, a fundamental unit in chemistry that enables chemists to count atoms, molecules, ions, and other entities in bulk samples.

To apply stoichiometry to balance a chemical equation, as shown in each step of the provided solutions, involves ensuring that the number of atoms of each element in the reactants side is equal to the number of atoms of the same element in the products side. This ensures adherence to the Law of Conservation of Mass. Additionally, stoichiometry allows chemists to predict the amounts of products that can be formed from given reactants, an essential step in predicting yields in reactions, as seen in the balancing of the decomposition of sodium peroxide into sodium hydroxide and oxygen gas.

Effective stoichiometry entails the use of coefficients to balance chemical equations. As advised, solving stoichiometric problems requires a systematic approach—writing down known information, identifying limiting reactants, and applying mole-to-mole conversions to determine the amounts of products expected from a reaction.

Classifying Reaction Types

Recognizing the type of chemical reaction is just as important as balancing the equation itself. Chemistry identifies several reaction types, including synthesis (combination), decomposition, single replacement, double replacement, combustion, and more.

Each reaction in the exercises illustrates different types of chemical changes:

• A synthesis reaction, where smaller molecules combine to form a larger one (e.g., the combination of CaO and H₂O to form Ca(OH)₂).
• Acid-base reactions, which are a subcategory of double displacement reactions, involve the transfer of protons between reactants (as seen with Al₂O₃ and H⁺ forming Al³⁺ and H₂O).
• A decomposition reaction occurs when a single compound breaks down into two or more elements or new compounds (Na₂O₂ decomposing into NaOH and O₂).

By identifying the reaction type, students can better predict the products and understand the changes at a molecular level. Moreover, appreciating the different reaction types also reinforces stoichiometry and equation balancing skills, as each reaction type follows specific patterns that once recognized, can make the balancing process more intuitive.