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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 \(\mathrm{N}_{2}\) ).

Short Answer

Expert verified
(a) Zn (s) + 2H鈦 (aq) 鈫 Zn虏鈦 (aq) + H鈧 (g) (b) S (s) + 6H鈦 (aq) + 6NO鈧冣伝 (aq) 鈫 SO鈧劼测伝 (aq) + 6NO鈧 (g) + 4H鈧侽 (l) (c) SO鈧 (g) + 2H鈦 (aq) + NO鈧冣伝 (aq) 鈫 SO鈧劼测伝 (aq) + H鈧侽 (l) + NO (g) (d) N鈧侶鈧 (g) + 6F鈧 (g) 鈫 4NF鈧 (g) (e) N鈧侶鈧 (aq) + 2CrO鈧劼测伝 (aq) + 8OH鈦 (aq) 鈫 2Cr(OH)鈧勨伝 (aq) + N鈧 (g)

Step by step solution

01

(a) Dilute nitric acid reacts with zinc metal with formation of nitrous oxide.

1. Write down the reaction: Zn (s) + HNO鈧 (aq) 鈫 Zn(NO鈧)鈧 (aq) + N鈧侽 (g) + H鈧侽 (l) 2. Balance the chemical equation: Zn (s) + 2HNO鈧 (aq) 鈫 Zn(NO鈧)鈧 (aq) + N鈧侽 (g) + H鈧侽 (l) 3. Write down the net ionic equation: Zn (s) + 2H鈦 (aq) 鈫 Zn虏鈦 (aq) + H鈧 (g)
02

(b) Concentrated nitric acid reacts with sulfur with formation of nitrogen dioxide.

1. Write down the reaction: S (s) + HNO鈧 (aq) 鈫 H鈧係O鈧 (aq) + NO鈧 (g) 2. Balance the chemical equation: S (s) + 6HNO鈧 (aq) 鈫 H鈧係O鈧 (aq) + 6NO鈧 (g) + 2H鈧侽 (l) 3. Write down the net ionic equation: S (s) + 6H鈦 (aq) + 6NO鈧冣伝 (aq) 鈫 SO鈧劼测伝 (aq) + 6NO鈧 (g) + 4H鈧侽 (l)
03

(c) Concentrated nitric acid oxidizes sulfur dioxide with formation of nitric oxide.

1. Write down the reaction: SO鈧 (g) + HNO鈧 (aq) 鈫 H鈧係O鈧 (aq) + NO (g) 2. Balance the chemical equation: 2SO鈧 (g) + 2HNO鈧 (aq) 鈫 2H鈧係O鈧 (aq) + 2NO (g) 3. Write down the net ionic equation: SO鈧 (g) + 2H鈦 (aq) + NO鈧冣伝 (aq) 鈫 SO鈧劼测伝 (aq) + H鈧侽 (l) + NO (g)
04

(d) Hydrazine is burned in excess fluorine gas, forming \(\mathrm{NF}_{3}\).

1. Write down the reaction: N鈧侶鈧 (g) + F鈧 (g) 鈫 2NF鈧 (g) 2. Balance the chemical equation: N鈧侶鈧 (g) + 6F鈧 (g) 鈫 4NF鈧 (g) 3. Write down the net ionic equation (There are no spectator ions in this case so the full equation is used): N鈧侶鈧 (g) + 6F鈧 (g) 鈫 4NF鈧 (g)
05

(e) Hydrazine reduces \(\mathrm{CrO}_{4}{ }^{2-}\) to \(\mathrm{Cr}(\mathrm{OH})_{4}{ }^{-}\)in base (hydrazine is oxidized to \(\mathrm{N}_{2}\)).

1. Write down the reaction: N鈧侶鈧 (aq) + CrO鈧劼测伝 (aq) + OH鈦 (aq) 鈫 Cr(OH)鈧勨伝 (aq) + N鈧 (g) 2. Balance the chemical equation: N鈧侶鈧 (aq) + 2CrO鈧劼测伝 (aq) + 8OH鈦 (aq) 鈫 2Cr(OH)鈧勨伝 (aq) + N鈧 (g) + 4H鈧侽 (l) 3. Write down the net ionic equation: N鈧侶鈧 (aq) + 2CrO鈧劼测伝 (aq) + 8OH鈦 (aq) 鈫 2Cr(OH)鈧勨伝 (aq) + N鈧 (g)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balancing Chemical Equations
Balancing chemical equations is a foundational skill in chemistry. It involves making sure that the number of atoms for each element is the same on both sides of the equation. This principle is grounded in the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. When you start balancing, begin by writing down the unbalanced equation.

Here are some tips to make the process easier:
  • Identify each component of the reaction, i.e., reactants and products.
  • Count the number of atoms of each element in the reactants and the products.
  • Adjust the coefficients of the reactants and products to achieve the same number of atoms for each element on both sides.
Remember, you cannot change the subscripts in the chemical formula as it would change the substance itself.

For example, in the process of balancing the reaction between zinc metal and dilute nitric acid, we started with Zn (s) + HNO鈧 (aq) 鈫 Zn(NO鈧)鈧 (aq) + N鈧侽 (g) + H鈧侽 (l). Upon balancing, the equation becomes Zn (s) + 2HNO鈧 (aq) 鈫 Zn(NO鈧)鈧 (aq) + N鈧侽 (g) + H鈧侽 (l). By doing this, we've ensured that all atoms are accounted for on both sides.
Oxidation-Reduction Reactions
Oxidation-reduction reactions, often referred to as redox reactions, are chemical processes where the oxidation state of atoms changes through the transfer of electrons. These reactions are split into two half-reactions: oxidation, where electrons are lost, and reduction, where electrons are gained.

Understanding redox reactions:
  • Oxidizing Agent: The substance that gains electrons (is reduced).
  • Reducing Agent: The substance that loses electrons (is oxidized).
  • Track the change in oxidation states to identify which element undergoes oxidation and which undergoes reduction.
For instance, when concentrated nitric acid reacts with sulfur, sulfur is oxidized to sulfur dioxide, a process where sulfur changes from a neutral oxidation state to a positive one.

The balanced ionic equation for this example highlights the role each substance plays: S (s) + 6H鈦 (aq) + 6NO鈧冣伝 (aq) 鈫 SO鈧劼测伝 (aq) + 6NO鈧 (g) + 4H鈧侽 (l). Understanding how each component contributes to electron movement is key in mastering redox reactions.
Chemical Reactions in Aqueous Solutions
Chemical reactions in aqueous solutions are reactions that occur with reactants dissolved in water. These solutions are common in a variety of chemical processes because water is a versatile solvent. The interactions can lead to precipitation, displacement, and redox reactions.

Some important aspects to consider:
  • Electrolytes: Substances that dissolve in water to form ions, contributing to electrical conductivity.
  • Net Ionic Equations: Show only the species that actually participate in the reaction, omitting the spectator ions.
  • Identifying strong acids, bases, and salts can often simplify predicting products and writing balanced equations.
For example, in reactions involving hydrazine reducing chromate, you might see a net ionic equation like N鈧侶鈧 (aq) + 2CrO鈧劼测伝 (aq) + 8OH鈦 (aq) 鈫 2Cr(OH)鈧勨伝 (aq) + N鈧 (g). Here, the emphasis is on the ions actively participating, providing a clearer view of the chemistry taking place in solution.

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Most popular questions from this chapter

What does hydrogen have in common with the halogens? Explain.

Explain the following observations: (a) For a given oxidation state, the acid strength of the oxyacid in aqueous solution decreases in the order chlorine \(>\) bromine \(>\) iodine. (b) \(\mathrm{Hy}\) drofluoric acid cannot be stored in glass bottles. (c) HI cannot be prepared by treating NaI with sulfuric acid. (d) The interhalogen \(\mathrm{ICl}_{3}\) is known, but \(\mathrm{BrCl}_{3}\) is not. Oxygen and the Other Group \(6 \mathrm{~A}\) Elements (Sections \(22.5\) and 22.6)

Write a balanced equation for each of the following reactions. (You may have to guess at one or more of the reaction products, but you should be able to make a reasonable guess, based on your study of this chapter.) (a) Hydrogen selenide can be prepared by reaction of an aqueous acid solution on aluminum selenide. (b) Sodium thiosulfate is used to remove excess \(\mathrm{Cl}_{2}\) from chlorine-bleached fabrics. The thiosulfate ion forms \(\mathrm{SO}_{4}{ }^{2-}\) and elemental sulfur, while \(\mathrm{Cl}_{2}\) is reduced to \(\mathrm{Cl}^{-}\). Nitrogen and the Other Group 5A Elements (Sections 22.7 and 22.8)

Write balanced equations for each of the following reactions. (a) When mercury(II) oxide is heated, it decomposes to form \(\mathrm{O}_{2}\) and mercury metal. (b) When copper(II) nitrate is heated strongly, it decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen. (c) Lead(II) sulfide, \(\mathrm{PbS}(s)\), reacts with ozone to form \(\mathrm{PbSO}_{4}(s)\) and \(\mathrm{O}_{2}(g)\). (d) When heated in air, \(\mathrm{ZnS}(s)\) is converted to \(\mathrm{ZnO}\). (e) Potassium peroxide reacts with \(\mathrm{CO}_{2}(g)\) to give potassium carbonate and \(\mathrm{O}_{2}\) (f) Oxygen is converted to ozone in the upper atmosphere.

Write a balanced equation for each of the following reactions: (a) hydrolysis of \(\mathrm{PCl}_{5}\) (b) dehydration of phosphoric acid (also called orthophosphoric acid) to form pyrophosphoric acid, (c) reaction of \(\mathrm{P}_{4} \mathrm{O}_{10}\) with water. Carbon, the Other Group 4A Elements, and Boron (Sections 22.9, 22.10, and 22.11)

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