Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 45

The following reactions do not occur in aqueous solutions. Balance their equations by the half-equation method, as suggested in Are You Wondering \(5-2\) (a) \(\mathrm{CH}_{4}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) \longrightarrow\) \(\mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{SO}_{2}(\mathrm{g}) \longrightarrow \mathrm{S}_{8}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \longrightarrow\) \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\)

Short Answer

Expert verified
The balanced equations are: (a): \(\mathrm{CH}_{4}(\mathrm{g})+2\mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+2\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), (b): \(8\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+8\mathrm{SO}_{2}(\mathrm{g}) \rightarrow \mathrm{S}_{8}(\mathrm{s})+8\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), (c): \(3\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+8\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+6\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\)

Step by step solution

01

Balancing Reaction (a)

For the reaction \(\mathrm{CH}_{4}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), start by balancing the carbon atoms, followed by the nitrogen atoms and finally the hydrogen atoms. This yields the balanced equation \(\mathrm{CH}_{4}(\mathrm{g})+2\mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+2\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).
02

Balancing Reaction (b)

For the reaction \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{SO}_{2}(\mathrm{g}) \rightarrow \mathrm{S}_{8}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), we first balance the sulfur atoms, then the hydrogen atoms, and finally the oxygen atoms. The balanced equation is thus \(8\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+8\mathrm{SO}_{2}(\mathrm{g}) \rightarrow \mathrm{S}_{8}(\mathrm{s})+8\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).
03

Balancing Reaction (c)

For the reaction \(\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\), the balancing starts with nitrogen, then hydrogen, and finally chlorine and oxygen. The balanced equation becomes \(3\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+8\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+6\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Iron (Fe) is obtained from rock that is extracted from open pit mines and then crushed. The process used to obtain the pure metal from the crushed rock produces solid waste, called tailings, which are stored in disposal areas near the mines. The tailings pose a serious environmental risk because they contain sulfides, such as pyrite ( \(\mathrm{FeS}_{2}\) ), which oxidize in air to produce metal ions and \(\mathrm{H}^{+}\) ions that can enter into surface water or ground water. The oxidation of \(\mathrm{FeS}_{2}\) to \(\mathrm{Fe}^{3+}\) is described by the unbalanced chemical equation below. \(\mathrm{FeS}_{2}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) \(\quad \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \quad(\text { not balanced })\) Thus, the oxidation of pyrite produces \(\mathrm{Fe}^{3+}\) and \(\mathrm{H}^{+}\) ions that can leach into surface or ground water. The leaching of \(\mathrm{H}^{+}\) ions causes the water to become very acidic. To prevent acidification of nearby ground or surface water, limestone \(\left(\mathrm{CaCO}_{3}\right)\) is added to the tailings to neutralize the \(\mathrm{H}^{+}\) ions: \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \underset{\mathrm{Ca}^{2+}}{\longrightarrow}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g})\) (a) Balance the equation above for the reaction of \(\mathrm{FeS}_{2}\) and \(\mathrm{O}_{2}\). [ Hint: Start with the half-equations \(\mathrm{FeS}_{2}(\mathrm{s}) \rightarrow\) \(\left.\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \text { and } \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(1) .\right]\) (b) What is the minimum amount of \(\mathrm{CaCO}_{3}(\mathrm{s})\) required, per kilogram of tailings, to prevent contamination if the tailings contain \(3 \%\) S by mass? Assume that all the sulfur in the tailings is in the form \(\mathrm{FeS}_{2}\).

Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), is a reducing agent that can be oxidized to different products, depending on the strength of the oxidizing agent and other conditions. By adding \(\mathrm{H}^{+}, \mathrm{H}_{2} \mathrm{O},\) and/or \(\mathrm{OH}^{-}\) as necessary, write redox equations to show the oxidation of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to (a) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) by \(\mathrm{I}_{2}\) (iodide ion is another product) (b) \(\mathrm{HSO}_{4}^{-}\) by \(\mathrm{Cl}_{2}\) (chloride ion is another product) (c) \(\mathrm{SO}_{4}^{2-}\) by \(\mathrm{OCl}^{-}\) in basic solution (chloride ion is another product)

In the half-reaction in which \(\mathrm{NpO}_{2}^{+}\) is converted to \(\mathrm{Np}^{4+},\) the number of electrons appearing in the half-equation is (a) \(1 ;(b) 2 ;(c) 3 ;\) (d) 4.

What volume of \(0.0962 \mathrm{M} \mathrm{NaOH}\) is required to exactly neutralize \(10.00 \mathrm{mL}\) of \(0.128 \mathrm{M} \mathrm{HCl} ?\)

Warfarin, \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4},\) is the active ingredient used in some anticoagulant medications. The amount of warfarin in a particular sample was determined as follows. A 13.96 g sample was first treated with an alkaline I_ solution to convert \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4}\) to \(\mathrm{CHI}_{3}\). This treatment gives one mole of \(\mathrm{CHI}_{3}\) for every mole of \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4}\) that was initially present in the sample. The iodine in \(\mathrm{CHI}_{3}\) is then precipitated as \(\mathrm{AgI}(\mathrm{s})\) by treatment with excess \(\mathrm{AgNO}_{3}(\mathrm{aq}):\) $$\begin{aligned} \mathrm{CHI}_{3}(\mathrm{aq})+3 \mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow & \longrightarrow 3 \mathrm{AgI}(\mathrm{s})+3 \mathrm{HNO}_{3}(\mathrm{aq}) &+\mathrm{CO}(\mathrm{g}) \end{aligned}$$ If \(0.1386 \mathrm{g}\) solid \(\mathrm{AgI}\) were obtained, then what is the percentage by mass of warfarin in the sample analyzed?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free