/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 177 Ammonia reacts with \(\mathrm{O}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 177

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(g)\) or \(\mathrm{NO}_{2}(g)\) according to these unbalanced equations: $$\begin{array}{l}\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\\\\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\end{array}$$ In a certain experiment 2.00 moles of \(\mathrm{NH}_{3}(g)\) and 10.00 moles of \(\mathbf{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, 6.75 moles of \(\mathbf{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

Short Answer

Expert verified
In the product mixture, there are \(0.752\) moles of \(\mathrm{NO}(g)\) formed.

Step by step solution

01

Balancing the reactions

First, we need to balance the given reactions: 1. \(\mathrm{NH}_{3}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g) + \mathrm{H}_{2} \mathrm{O}(g)\) 4\(\mathrm{NH}_{3}(g) + 5\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{NO}(g) + 6\mathrm{H}_{2} \mathrm{O}(g)\) 2. \(\mathrm{NH}_{3}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(g)\) 4\(\mathrm{NH}_{3}(g) + 7\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{NO}_{2}(g) + 6\mathrm{H}_{2} \mathrm{O}(g)\)
02

Determine the moles of \(\mathrm{O}_{2}\) used in each reaction

We know that 10.00 moles of \(\mathrm{O}_{2}(g)\) were initially present, and after the reaction, 6.75 moles remain. This means that: Total moles of \(\mathrm{O}_{2}\) used = Initial moles of \(\mathrm{O}_{2}\) - Remaining moles of \(\mathrm{O}_{2}\) Total moles of \(\mathrm{O}_{2}\) used = 10.00 moles - 6.75 moles = 3.25 moles Now, let's consider that \(y\) moles of \(\mathrm{O}_{2}(g)\) are used in the first reaction and \(x\) moles of \(\mathrm{O}_{2}(g)\) are used in the second reaction. Then, we can write the following equation: \(x + y = 3.25\)
03

Calculate the moles of \(\mathrm{NO}(g)\) formed in the reaction

In the balanced reaction 1, 4 moles of \(\mathrm{NH}_{3}(g)\) react with 5 moles of \(\mathrm{O}_{2}(g)\) to produce 4 moles of \(\mathrm{NO}(g)\). From this information, we can get a proportionality constant, which we can use to calculate the moles of \(\mathrm{NO}(g)\) formed when a specific amount of \(\mathrm{O}_{2}(g)\) is consumed in the reaction. In reaction 1, we have: \(\frac{\text{Moles of \)\mathrm{NO}\( formed}}{\text{Moles of \)\mathrm{O}_{2}\( consumed}} = \frac{4}{5}\) Now, we know that 2.00 moles of \(\mathrm{NH}_{3}(g)\) is completely consumed in both reactions. We can write a similar relation of moles for reaction 2: \(\frac{\text{Moles of \)\mathrm{NH}_{3}\( consumed in reaction 1}}{\text{Moles of \)\mathrm{NH}_{3}\( consumed in reaction 2}} = \frac{4}{4}\) \(\frac{2 - y/5}{x/7} = 1\) Now we have two equations: 1. \(x + y = 3.25\) 2. \(2 - y/5 = x/7\) By solving these two equations simultaneously for \(x\) and \(y\), we get: \(x = 2.31\) (moles of \(\mathrm{O}_2\) consumed in the second reaction) \(y = 0.94\) (moles of \(\mathrm{O}_2\) consumed in the first reaction) Now we can find the moles of \(\mathrm{NO}(g)\) formed in the first reaction: Moles of \(\mathrm{NO}(g)\) = \(\frac{\text{4 moles}}{\text{5 moles of \)\mathrm{O}_{2}\( consumed}} \times \text{0.94 moles of \)\mathrm{O}_{2}\( consumed}\) Moles of \(\mathrm{NO}(g)\) = \(\frac{4}{5} \times 0.94\) Moles of \(\mathrm{NO}(g)\) formed = \(\boxed{0.752}\)

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Stoichiometry
Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. It allows chemists to predict how much of a substance will be consumed or produced in a given reaction. Understanding stoichiometry involves using the molar ratios found in balanced chemical equations. These ratios are essential to calculating the exact amounts of chemicals needed or produced.

For example, in our exercise, we determine how many moles of ammonia (\(\text{NH}_3\)) react with oxygen (\(\text{O}_2\)) to form nitrogen monoxide (\(\text{NO}\)) or nitrogen dioxide (\(\text{NO}_2\)). By utilizing stoichiometry, we can calculate that 3.25 moles of \(\text{O}_2\) were used in the reaction, helping us find how much \(\text{NO}\) was produced. This concept empowers us to precisely predict and control chemical processes, making stoichiometry an invaluable tool in chemistry.
Chemical Equations
Chemical equations are symbolic representations of chemical reactions. They showcase the substances involved, both reactants and products, and their relative quantities. To accurately convey these transformations, chemical equations must be balanced. Balancing ensures that the number of atoms of each element is the same on both sides of the equation, adhering to the Law of Conservation of Mass.

In the given problem, ammonia reacts with oxygen to create nitrogen oxides and water. The equations provided initially were unbalanced:
  • \(\text{NH}_3(g) + \text{O}_2(g) \longrightarrow \text{NO}(g) + \text{H}_2\text{O}(g)\)
  • \(\text{NH}_3(g) + \text{O}_2(g) \longrightarrow \text{NO}_2(g) + \text{H}_2\text{O}(g)\)
By balancing these, we ensure that chemical identities and mass are conserved. Balancing helps us derive the correct stoichiometric relationships necessary for calculations.
Reaction Balancing
Reaction balancing is the process of equalizing the number of atoms for each element on both sides of a chemical equation. It鈥檚 a fundamental requirement in chemistry to maintain the law of conservation of mass. The balanced form of a chemical equation gives insight into the stoichiometric relationships between reactants and products.

In our problem, the reactions were balanced as follows:
  • \(4\text{NH}_3(g) + 5\text{O}_2(g) \longrightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g)\)
  • \(4\text{NH}_3(g) + 7\text{O}_2(g) \longrightarrow 4\text{NO}_2(g) + 6\text{H}_2\text{O}(g)\)
Balancing these equations allowed us to apply the stoichiometric ratios effectively. For instance, knowing that 4 moles of \(\text{NH}_3\) yield 4 moles of \(\text{NO}\) while consuming 5 moles of \(\text{O}_2\), we can correctly deduce the amount of \(\text{NO}\) formed, as demonstrated in the exercise where we calculated 0.752 moles of \(\text{NO}\). Reaction balancing transforms a theoretical equation into a practical framework for conducting and understanding chemical reactions.

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

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced \(0.213 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0310 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that 0.103 g of the compound produces \(0.0230 \mathrm{g} \mathrm{NH}_{3}\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

The aspirin substitute. acetaminophen \(\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{N}\right),\) is produced by the following three-step synthesis: I. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{N}(s)+3 \mathrm{H}_{2}(g)+\mathrm{HCl}(a q) \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) II. \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+\mathrm{NaOH}(a q) \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCl}(a q)\) III. \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}(l) \longrightarrow\) \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{N}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(l)\) The first two reactions have percent yields of \(87 \%\) and \(98 \%\) by mass, respectively. The overall reaction yields 3 moles of acetaminophen product for every 4 moles of \(C_{6} H_{5} O_{3} N\) reacted. a. What is the percent yield by mass for the overall process? b. What is the percent yield by mass of Step III?

Consider the following unbalanced equation: $$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q)$$ What masses of calcium sulfate and phosphoric acid can be produced from the reaction of \(1.0 \mathrm{kg}\) calcium phosphate with \(1.0 \mathrm{kg}\) concentrated sulfuric acid \(\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\text { by mass)? }\right.\)

A gas contains a mixture of \(\mathrm{NH}_{3}(g)\) and \(\mathrm{N}_{2} \mathrm{H}_{4}(g),\) both of which react with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{NO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) The gaseous mixture (with an initial mass of \(61.00 \mathrm{g}\) ) is reacted with 10.00 moles \(\mathrm{O}_{2}\), and after the reaction is complete, 4.062 moles of \(\mathrm{O}_{2}\) remains. Calculate the mass percent of \(\mathrm{N}_{2} \mathrm{H}_{4}(g)\) in the original gaseous mixture.

Boron consists of two isotopes, \(^{10} \mathbf{B}\) and \(^{11} \mathbf{B}\). Chlorine also has two isotopes, \(^{35} \mathrm{Cl}\) and \(^{37} \mathrm{Cl}\). Consider the mass spectrum of \(\mathrm{BCl}_{3}\) How many peaks would be present, and what approximate mass would each peak correspond to in the \(\mathrm{BCl}_{3}\) mass spectrum?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

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.