Principles of General Chemistry
Principles of General Chemistry
3rd Edition
ISBN: 9780073402697
Author: SILBERBERG, Martin S.
Publisher: McGraw-Hill College
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 3, Problem 3.81P

(a)

Interpretation Introduction

Interpretation: The balanced equation corresponding to the statement when hydrogen sulfide forms sulfur dioxide and water vapor on reaction with oxygen should be written.

Concept introduction: In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

(a)

Expert Solution
Check Mark

Answer to Problem 3.81P

The balanced equation is written as follows:

  2H2S(s)+3O2(g)2SO2(g)+2H2O(g)

Explanation of Solution

The formula ofgallium oxide is H2S .The unbalanced equation that needs to be balanced is as follows:

  H2S(s)+O2(g)SO2(g)+H2O(g)

Since, left side has 2 O while right side has 3 O so first the stoichiometric coefficient of O2 on the left is made 3/2.

  H2S(s)+32O2(g)SO2(g)+H2O(g)

Multiple the entire equation by 2.

  2H2S(s)+3O2(g)2SO2(g)+2H2O(g)

Since above equation has equal atoms on right and left side so the equation is now balanced.

(b)

Interpretation Introduction

Interpretation: The balanced equation corresponding to statement when potassium chlorate is heated to form two potassium chloride and potassium perchlorate should be written.

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

(b)

Expert Solution
Check Mark

Answer to Problem 3.81P

The balanced equation is written as follows:

  4KClO3(s)Δ3KClO4(s)+KCl(s)

Explanation of Solution

The formula ofpotassium chlorate and potassium perchlorate are KClO3 and KClO4 . The unbalanced equation that needs to be balanced is as follows:

  KClO3(s)ΔKClO4(s)+KCl(s)

Since, left side has 3 O while right side has 4 O so place 4 as the coefficient of KClO3 and 3 as coefficient of KClO4 so as to make equal O on both side.

  4KClO3(s)Δ3KClO4(s)+KCl(s)

Since above equation has equal atoms on each side so the equation is now balanced.

(c)

Interpretation Introduction

Interpretation: The balanced equation corresponding to the statement that when hydrogen gas is passed on iron(III) oxide to form iron metal and water vapor formshould be written.

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

(c)

Expert Solution
Check Mark

Answer to Problem 3.81P

The balanced equation is written as follows:

  3H2(g)+Fe3O3(s)3H2O(g)+2Fe(s)

Explanation of Solution

The formula ofcalcium chloride, sodium phosphate are CaCl2 and Na3PO4 respectively while the formula ofcalcium phosphate and sodium chloride are Ca3( PO4)2 and NaCl respectively.

The unbalanced equation that needs to be balanced is as follows:

  H2(g)+Fe3O3(s)H2O(g)+Fe(s)

Since, left side has 3 O while right side has 4 O so place 4 as the coefficient of KClO3 and 3 as coefficient of KClO4 so as to make equal O on both side.

  H2(g)+Fe3O3(s)3H2O(g)+Fe(s)

Place 3 as the coefficient of H2 so as to make equal 6 H on each side.

  3H2(g)+Fe3O3(s)3H2O(g)+Fe(s)

Place 2 as the coefficient of Fe so as to make 2 Fe on each side.

  3H2(g)+Fe3O3(s)3H2O(g)+2Fe(s)

Since above equation has equal atoms on right and left side so the equation is now balanced.

(d)

Interpretation Introduction

Interpretation: The balanced equation corresponding to the statement that when ethane undergoes combustion should be written.

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

(d)

Expert Solution
Check Mark

Answer to Problem 3.81P

The balanced equation is written as follows:

  2C2H6(g)+7O2(g)6H2O(g)+4CO2(g)

Explanation of Solution

The unbalanced equation that needs to be balanced is as follows:

  C2H6(g)+O2(g)H2O(g)+CO2(g)

Place 3 as coefficient of H2O and 2 as coefficient of CO2 so as to make equal H and C on each side.

  C2H6(g)+O2(g)3H2O(g)+2CO2(g)

Since, left side has 2 O while right side has 7 O so place 7/2 as coefficient of O2 to make equal O on both side.

  C2H6(g)+72O2(g)3H2O(g)+2CO2(g)

Multiple the entire equation by 2.

  2C2H6(g)+7O2(g)6H2O(g)+4CO2(g)

Since above equation has equal atoms on right and left side so the equation is now balanced.

(e)

Interpretation Introduction

Interpretation: The balanced equation corresponding to the statement hen iron(II) chloride is converted to iron(III) ) fluoride should be written.

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

(e)

Expert Solution
Check Mark

Answer to Problem 3.81P

The balanced equation is written as follows:

  2FeCl2(s)+2ClF3(g)2FeF3(s)+3Cl2(g)

Explanation of Solution

The unbalanced equation when iron(II) chloride is converted to iron(III) ) fluoride is as follows:

  FeCl2(s)+ClF3(g)FeF3(s)+Cl2(g)

Place 3 as coefficient of ClF3 to make equal Cl on each side.

  FeCl2(s)+2ClF3(g)FeF3(s)+2Cl2(g)

Place 2 as coefficient of FeF3 to make equal F on each side.

  FeCl2(s)+2ClF3(g)2FeF3(s)+2Cl2(g)

Place 2 as coefficient of FeCl2 to make equal Fe on each side and change coefficient of Cl2 from 2 to 3.

  2FeCl2(s)+2ClF3(g)2FeF3(s)+3Cl2(g)

Since above equation has equal atoms on right and left side so the equation is now balanced.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
3. Provide all the steps and reagents for this synthesis. OH
Steps and explanation
Steps and explanations please.

Chapter 3 Solutions

Principles of General Chemistry

Ch. 3 - Prob. 3.11PCh. 3 - Calculate each of the following quantities: (a)...Ch. 3 - Prob. 3.13PCh. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Calculate each of the following quantities: (a)...Ch. 3 - Calculate each of the following: Mass % of H in...Ch. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Prob. 3.20PCh. 3 - Prob. 3.21PCh. 3 - Prob. 3.22PCh. 3 - Prob. 3.23PCh. 3 - Which of the following sets of information allows...Ch. 3 - What is the empirical formula and empirical...Ch. 3 - Prob. 3.26PCh. 3 - Prob. 3.27PCh. 3 - Prob. 3.28PCh. 3 - Prob. 3.29PCh. 3 - Prob. 3.30PCh. 3 - Prob. 3.31PCh. 3 - Prob. 3.32PCh. 3 - Cortisol (m=362.47g/mol) is a steroid hormone...Ch. 3 - Prob. 3.34PCh. 3 - Prob. 3.35PCh. 3 - Prob. 3.36PCh. 3 - Write balanced equations for each of the following...Ch. 3 - Write balanced equations for each of the following...Ch. 3 - Prob. 3.39PCh. 3 - Prob. 3.40PCh. 3 - Prob. 3.41PCh. 3 - Potassium nitrate decomposes on heating, producing...Ch. 3 - Prob. 3.43PCh. 3 - Calculate the mass of each product formed when...Ch. 3 - Prob. 3.45PCh. 3 - Prob. 3.46PCh. 3 - Prob. 3.47PCh. 3 - Many metals react with oxygen gas to form the...Ch. 3 - Prob. 3.49PCh. 3 - Calculate the maximum numbers of moles and grams...Ch. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54PCh. 3 - Prob. 3.55PCh. 3 - Prob. 3.56PCh. 3 - Prob. 3.57PCh. 3 - Prob. 3.58PCh. 3 - Prob. 3.59PCh. 3 - Prob. 3.60PCh. 3 - Prob. 3.61PCh. 3 - Prob. 3.62PCh. 3 - Prob. 3.63PCh. 3 - Prob. 3.64PCh. 3 - Prob. 3.65PCh. 3 - Six different aqueous solutions (with solvent...Ch. 3 - Prob. 3.67PCh. 3 - Prob. 3.68PCh. 3 - Prob. 3.69PCh. 3 - Calculate each of the following quantities: (a)...Ch. 3 - Prob. 3.71PCh. 3 - Prob. 3.72PCh. 3 - Prob. 3.73PCh. 3 - Prob. 3.74PCh. 3 - Prob. 3.75PCh. 3 - Prob. 3.76PCh. 3 - Prob. 3.77PCh. 3 - Prob. 3.78PCh. 3 - Prob. 3.79PCh. 3 - Prob. 3.80PCh. 3 - Prob. 3.81PCh. 3 - Prob. 3.82PCh. 3 - Prob. 3.83PCh. 3 - Prob. 3.84PCh. 3 - Prob. 3.85PCh. 3 - Seawater is approximately 4.0% by mass dissolved...Ch. 3 - Is each of the following statements true or false?...Ch. 3 - Prob. 3.88PCh. 3 - In each pair, choose the larger of the indicated...Ch. 3 - Prob. 3.90PCh. 3 - Prob. 3.91PCh. 3 - Assuming that the volumes are additive, what is...Ch. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - Hydrocarbon mixtures are used as fuels, (a) How...Ch. 3 - Prob. 3.96PCh. 3 - Prob. 3.97PCh. 3 - Prob. 3.98PCh. 3 - Prob. 3.99PCh. 3 - Write a balanced equation for the reaction...Ch. 3 - Prob. 3.101PCh. 3 - Citric acid (right) is concentrated in citrus...Ch. 3 - Prob. 3.103PCh. 3 - Prob. 3.104PCh. 3 - Prob. 3.105PCh. 3 - Prob. 3.106PCh. 3 - Aspirin (acetylsalicylic acid, C9H8O4 ) is made by...Ch. 3 - Prob. 3.108PCh. 3 - Prob. 3.109PCh. 3 - Prob. 3.110PCh. 3 - High-temperature superconducting oxides hold great...
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Text book image
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Living By Chemistry: First Edition Textbook
Chemistry
ISBN:9781559539418
Author:Angelica Stacy
Publisher:MAC HIGHER
Types of Matter: Elements, Compounds and Mixtures; Author: Professor Dave Explains;https://www.youtube.com/watch?v=dggHWvFJ8Xs;License: Standard YouTube License, CC-BY