Problem 87 - Chemistry The Molecular Nature of Matter

Question

Complete and balance the following equations. For each, write the molecular, ionic, and net ionic equations. (All of the products are soluble in water.) (a) \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HNO}_{3}(a q) \longrightarrow\) (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{HCl}(a q) \longrightarrow\) (c) \(\mathrm{Zn}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\)

Step-by-Step Solution

Answer

Part A: \(\mathrm{Ca(OH)_{2} (aq) + 2HNO_{3} (aq) \longrightarrow Ca(NO_{3})_{2} (aq) + 2H_{2}O (l)}\), net ionic: \(2OH^{-} (aq) + 2H^{+} (aq) \longrightarrow 2H_{2}O (l)\). Part B: \(\mathrm{Al_{2}O_{3} (s) + 6HCl (aq) \longrightarrow 2AlCl_{3} (aq) + 3H_{2}O (l)}\), net ionic: \(\mathrm{Al_{2}O_{3} (s) + 6H^{+} (aq) \longrightarrow 2Al^{3+} (aq) + 3H_{2}O (l)}\). Part C: \(\mathrm{Zn(OH)_{2} (s) + H_{2}SO_{4} (aq) \longrightarrow ZnSO_{4} (aq) + 2H_{2}O (l)}\), net ionic: \(\mathrm{Zn(OH)_{2} (s) + 2H^{+} (aq) \longrightarrow Zn^{2+} (aq) + 2H_{2}O (l)}\).

Step 1: Part A - Balancing the Molecular Equation

Write the unbalanced molecular equation: \(\mathrm{Ca(OH)_{2} (aq) + HNO_{3} (aq) \longrightarrow Ca(NO_{3})_{2} (aq) + H_{2}O (l)}\). Balance the equation: For every 1 mole of \(Ca(OH)_{2}\), 2 moles of \(HNO_{3}\) are needed to produce 1 mole of \(Ca(NO_{3})_{2}\) and 2 moles of \(H_{2}O\). The balanced molecular equation is \(\mathrm{Ca(OH)_{2} (aq) + 2HNO_{3} (aq) \longrightarrow Ca(NO_{3})_{2} (aq) + 2H_{2}O (l)}\).

Step 2: Part A - Writing the Ionic Equation

Next, write the ionic equation by showing strong electrolytes as they exist in solution - ions: \(\mathrm{Ca^{2+} (aq) + 2OH^{-} (aq) + 2H^{+} (aq) + 2NO_{3}^{-} (aq) \longrightarrow Ca^{2+} (aq) + 2NO_{3}^{-} (aq) + 2H_{2}O (l)}\).

Step 3: Part A - Writing the Net Ionic Equation

Cancel the spectator ions that appear on both sides of the ionic equation to get the net ionic equation: The spectator ions are \(\mathrm{Ca^{2+} (aq)}\) and \(\mathrm{2NO_{3}^{-} (aq)}\). The resulting net ionic equation is \(\mathrm{2OH^{-} (aq) + 2H^{+} (aq) \longrightarrow 2H_{2}O (l)}\).

Step 4: Part B - Balancing the Molecular Equation

First, write the unbalanced molecular equation: \(\mathrm{Al_{2}O_{3} (s) + HCl (aq) \longrightarrow AlCl_{3} (aq) + H_{2}O (l)}\). Balance the equation: For every mole of \(Al_{2}O_{3}\), 6 moles of \(HCl\) are needed to produce 2 moles of \(AlCl_{3}\) and 3 moles of \(H_{2}O\). The balanced molecular equation is \(\mathrm{Al_{2}O_{3} (s) + 6HCl (aq) \longrightarrow 2AlCl_{3} (aq) + 3H_{2}O (l)}\).

Step 5: Part B - Writing the Ionic Equation

Because \(Al_{2}O_{3}\) does not dissociate (it's a solid), leave it intact, while showing the other strong electrolytes as ions: \(\mathrm{Al_{2}O_{3} (s) + 6H^{+} (aq) + 6Cl^{-} (aq) \longrightarrow 2Al^{3+} (aq) + 6Cl^{-} (aq) + 3H_{2}O (l)}\).

Step 6: Part B - Writing the Net Ionic Equation

The chloride ions are spectator ions and can be canceled. The net ionic equation is \(\mathrm{Al_{2}O_{3} (s) + 6H^{+} (aq) \longrightarrow 2Al^{3+} (aq) + 3H_{2}O (l)}\).

Step 7: Part C - Balancing the Molecular Equation

Begin with an unbalanced molecular equation: \(\mathrm{Zn(OH)_{2} (s) + H_{2}SO_{4} (aq) \longrightarrow ZnSO_{4} (aq) + H_{2}O (l)}\). Balancing gives: For every mole of \(Zn(OH)_{2}\), 1 mole of \(H_{2}SO_{4}\) is needed, resulting in 1 mole of \(ZnSO_{4}\) and 2 moles of \(H_{2}O\). The balanced molecular equation is \(\mathrm{Zn(OH)_{2} (s) + H_{2}SO_{4} (aq) \longrightarrow ZnSO_{4} (aq) + 2H_{2}O (l)}\).

Step 8: Part C - Writing the Ionic Equation

Write \(Zn(OH)_{2}\) as it is (a solid), and the strong electrolytes as ions: \(\mathrm{Zn(OH)_{2} (s) + 2H^{+} (aq) + SO_{4}^{2-} (aq) \longrightarrow Zn^{2+} (aq) + SO_{4}^{2-} (aq) + 2H_{2}O (l)}\).

Step 9: Part C - Writing the Net Ionic Equation

The sulfate ions are spectator ions and can be canceled. The net ionic equation is \(\mathrm{Zn(OH)_{2} (s) + 2H^{+} (aq) \longrightarrow Zn^{2+} (aq) + 2H_{2}O (l)}\).

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, ensuring that the same number of atoms for each element is present on both sides of the equation. It is a reflection of the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.

To balance an equation, first write down the unbalanced equation. Then, adjust the coefficients — the numbers before the chemical formulas — to get the same number of atoms of each element on both sides. For complex equations, it might be helpful to start by balancing elements that appear in the fewest compounds and save hydrogen and oxygen for last.

Take, for example, the reaction between calcium hydroxide and nitric acid. The balanced molecular equation is \(\mathrm{Ca(OH)_2 (aq) + 2HNO_3 (aq) \longrightarrow Ca(NO_3)_2 (aq) + 2H_2O (l)}\), where the coefficient '2' in front of \(HNO_3\) ensures that there are equal numbers of hydrogen and nitrate ions on both sides of the equation.

Molecular Equations

Molecular equations show the complete formulas of all reactants and products without indicating their ionic character. They represent the overall reaction without detailing the specific ionic species present in the solution. Molecular equations give a clear picture of the compounds involved in the reaction before they dissociate into ions.

The molecular form of the equation for the reaction of aluminum oxide with hydrochloric acid is \(\mathrm{Al_2O_3 (s) + 6HCl (aq) \longrightarrow 2AlCl_3 (aq) + 3H_2O (l)}\). In this molecular view, the substances are represented as if they exist as molecules, with no indication of ionization in solution.

Ionic Equations

Ionic equations are a more detailed representation of an aqueous chemical reaction that show the ionic species that are present in the reaction. These types of equations display strong electrolytes as they exist in solution — as separated ions — instead of compounds. By doing so, ionic equations acknowledge the actual forms of the reactants and products when dissolved in water.

For instance, the ionic equation for the neutralization reaction between calcium hydroxide and nitric acid is expressed as \(\mathrm{Ca^{2+} (aq) + 2OH^{-} (aq) + 2H^{+} (aq) + 2NO_3^{-} (aq) \longrightarrow Ca^{2+} (aq) + 2NO_3^{-} (aq) + 2H_2O (l)}\). This equation shows how each soluble substance separates into its constituent ions.

Spectator Ions

Spectator ions are ions in an ionic equation that do not participate in the chemical reaction and remain unchanged on both the reactant and product sides of the equation. They are essentially bystanders in the reaction, and identifying them helps to simplify and focus on the actual chemical change taking place.

When writing net ionic equations, the spectator ions are omitted to reveal the essence of the chemical reaction. For example, in the reaction between zinc hydroxide and sulfuric acid, the sulfate \(\mathrm{SO_4^{2-}}\) ions are spectator ions. They appear on both sides of the full ionic equation but are not included in the net ionic equation, which is \(\mathrm{Zn(OH)_2 (s) + 2H^{+} (aq) \longrightarrow Zn^{2+} (aq) + 2H_2O (l)}\), thus highlighting the direct interaction between zinc hydroxide and hydrogen ions.