Problem 54 - Chemistry The Central Science

Question

Write a balanced net ionic equation for each of the following reactions: (a) Dilute nitric acid reacts with zinc metal with formation of nitrous oxide. (b) Concentrated nitric acid reacts with sulfur with formation of nitrogen dioxide. (c) Concentrated nitric acid oxidizes sulfur dioxide with formation of nitric oxide. (d) Hydrazine is burned in excess fluorine gas, forming \(\mathrm{NF}_{3}\) . (e) Hydrazine reduces \(\mathrm{CrO}_{4}^{2-}\) to \(\mathrm{Cr}(\mathrm{OH})_{4}^{-}\) in base (hydrazine is oxidized to \(\mathrm{N}_{2} )\) .

Step-by-Step Solution

Answer

(a) \(Zn(s) + 4HNO_3(aq) \rightarrow N_2O(g) + 2NO_2(g) + Zn(NO_3)_2(aq) + 2H_2O(l)\) (b) \(S(s) + 6HNO_3(aq) \rightarrow 6NO_2(g) + H_2SO_4(aq) + 2H_2O(l)\) (c) \(2NO_3^-(aq) + 4H^+(aq) + SO_2(g) \rightarrow 2NO(g) + SO_4^{2-}(aq) + 2H_2O(l)\) (d) \(N_2H_4(l) + 6F_2(g) \rightarrow 4NF_3(g) + 2H_2(g)\) (e) \(N_2H_4(aq) + CrO_4^{2-}(aq) + 4OH^-(aq) \rightarrow N_2(g) + Cr(OH)_4^-(aq) + 4H_2O(l)\)

Step 1: Write the balanced chemical equation

: \(Zn(s) + 4HNO_3(aq) \rightarrow N_2O(g) + 2NO_2(g) + Zn(NO_3)_2(aq) + 2H_2O(l)\)

Step 2: Identify all the species and spectator ions

: All species in the reaction are included. There are no spectator ions in this reaction.

Step 3: Write the net ionic equation

: The net ionic equation is the same as the balanced chemical equation in this case. \(Zn(s) + 4HNO_3(aq) \rightarrow N_2O(g) + 2NO_2(g) + Zn(NO_3)_2(aq) + 2H_2O(l)\) #b) Concentrated nitric acid reacts with sulfur with formation of nitrogen dioxide.#

Step 1: Write the balanced chemical equation

: \(S(s) + 6HNO_3(aq) \rightarrow 6NO_2(g) + H_2SO_4(aq) + 2H_2O(l)\)

Step 2: Identify all the species and spectator ions

: All species in the reaction are included. There are no spectator ions in this reaction.

Step 3: Write the net ionic equation

: The net ionic equation is the same as the balanced chemical equation in this case. \(S(s) + 6HNO_3(aq) \rightarrow 6NO_2(g) + H_2SO_4(aq) + 2H_2O(l)\) #c) Concentrated nitric acid oxidizes sulfur dioxide with formation of nitric oxide.#

Step 1: Write the balanced chemical equation

: \(2NO_3^-(aq) + 4H^+(aq) + SO_2(g) \rightarrow 2NO(g) + SO_4^{2-}(aq) + 2H_2O(l)\)

Step 2: Identify all the species and spectator ions

: All species in the reaction are included, and there are no spectator ions.

Step 3: Write the net ionic equation

: The net ionic equation is the same as the balanced chemical equation in this case. \(2NO_3^-(aq) + 4H^+(aq) + SO_2(g) \rightarrow 2NO(g) + SO_4^{2-}(aq) + 2H_2O(l)\) #d) Hydrazine is burned in excess fluorine gas, forming NF3.#

Step 1: Write the balanced chemical equation

: \(N_2H_4(l) + 6F_2(g) \rightarrow 4NF_3(g) + 2H_2(g)\)

Step 2: Identify all the species and spectator ions

: All species in the reaction are included, and there are no spectator ions.

Step 3: Write the net ionic equation

: The net ionic equation is the same as the balanced chemical equation in this case. \(N_2H_4(l) + 6F_2(g) \rightarrow 4NF_3(g) + 2H_2(g)\) #e) Hydrazine reduces CrO42- to Cr(OH)4- in base (hydrazine is oxidized to N2).#

Step 1: Write the balanced chemical equation

: \(N_2H_4(aq) + CrO_4^{2-}(aq) + 4OH^-(aq) \rightarrow N_2(g) + Cr(OH)_4^-(aq) + 4H_2O(l)\)

Step 2: Identify all the species and spectator ions

: All species in the reaction are included, and there are no spectator ions.

Step 3: Write the net ionic equation

: The net ionic equation is the same as the balanced chemical equation in this case. \(N_2H_4(aq) + CrO_4^{2-}(aq) + 4OH^-(aq) \rightarrow N_2(g) + Cr(OH)_4^-(aq) + 4H_2O(l)\) Finally, we have all the balanced net ionic equations for each given reaction.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is critical to understanding chemistry. It ensures that the law of conservation of mass is followed. This law states that matter cannot be created or destroyed, merely changed from one form into another. In a balanced equation, the number of atoms for each element is the same on the reactant side as on the product side.

In practice, balancing chemical equations involves adjusting the coefficients (the numbers before the chemical formulas) to make the number of atoms for each element equal on both sides. It’s important to note that balancing an equation does not change the actual chemistry of the reaction; it simply assures that the equation accurately represents that chemistry. For example, in the reaction between zinc and nitric acid, the balanced chemical equation is:
\[Zn(s) + 4HNO_3(aq) \rightarrow N_2O(g) + 2NO_2(g) + Zn(NO_3)_2(aq) + 2H_2O(l)\].

Each side of the equation has one zinc atom, four nitrogen atoms, and eight oxygen atoms, satisfying the law of conservation of mass.

Reactivity of Zinc with Nitric Acid

The reactivity of zinc with nitric acid exemplifies a redox reaction where zinc undergoes oxidation and nitric acid is reduced. This interaction between a metal and an acid results in the evolution of gas; in the case of zinc and dilute nitric acid, nitrous oxide is produced. Reactivity can depend on the concentration of the acid. With dilute nitric acid, zinc reacts to form zinc nitrate, nitrous oxide, and water, as indicated in the equation from the solution steps.

It’s interesting to note that different reactions occur if the concentration of nitric acid changes. For example, concentrated nitric acid will result in the formation of nitrogen dioxide instead of nitrous oxide. This shows that concentration can significantly change the products of chemical reactions.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, involve the transfer of electrons from one substance to another. The substance that gives away electrons is oxidized, while the substance that gains electrons is reduced. When we balance redox equations, we must account for both the mass and charge to ensure the equation is balanced.

Oxidation numbers can help us identify which species are oxidized and which are reduced. For example, in the reaction of sulfur dioxide with concentrated nitric acid, sulfur dioxide is oxidized to sulfate (\(SO_2 \rightarrow SO_4^{2-}\)), and nitric acid is reduced to nitric oxide (\(NO_3^- \rightarrow NO\)). Each substance either gains or loses electrons, showing the redox nature of the chemical reaction.

Reactions of Nitric Acid

Nitric acid is a versatile and strong oxidizing agent, capable of participating in multiple reaction types. It reacts differently depending on its concentration and the type of substance it encounters. As seen in the textbook exercises, nitric acid can produce a variety of nitrogen oxides, such as nitrous oxide (\(N_2O\)) with zinc, nitrogen dioxide (\(NO_2\)) with sulfur, and nitric oxide (\(NO\)) with sulfur dioxide. The reactivity extends to different chemical environments, showing how nitric acid can act differently with metals, non-metals, and even within aqueous solutions. Understanding these reactions is crucial for students studying chemistry, environmental science, and many industrial processes.

It is key to remember that while these reactions produce different nitrogen oxides, they all follow the same rules of chemical reactions, such as the conservation of mass and charge, principles of balancing equations, and the ability to drive redox processes.