Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4, Problem 4.132AP

(a)

Interpretation Introduction

Interpretation: The questions corresponding to acidic mine drainage are to be answered.

Concept introduction: The number of given moles of solute divided by a liter of solution is termed as molarity. It can be given by the formula,

Molarity=Numberofmolesofsolute(n)Volumeofsolution(V)inL

To determine: The molarity of iron and zinc in the drainage.

(a)

Expert Solution
Check Mark

Answer to Problem 4.132AP

Solution

The molarity of iron is 1.43M_ and the molarity of zinc in the drainage is 9.17×10-2M_ .

Explanation of Solution

Explanation

The mass (g) of iron dissolved in one liter of drainage is 80.0g .

The molar mass of iron (Fe) is 55.85g/mol .

The mass (g) of zinc dissolved in one liter of drainage is 6.0g .

The molar mass of iron (Fe) is 65.41g/mol .

The volume of the drainage is 1L .

Firstly, the moles of iron and zinc are calculated that will be used in the formula of molarity further.

The moles of iron (Fe) and zinc (Zn) are calculated by the formula,

Moles=Mass(g)Molarmass

Substitute the values of mass and molar mass of iron (Fe) in the above formula as,

Molesofiron(Fe)=80.0g55.85g×1molFe=1.43molFe

Substitute the values of mass and molar mass of zinc (Zn) in the above formula as,

Molesofzinc(Zn)=6.0g65.41g×1molZn=9.17×102molZn

The molarity of iron and zinc in the drainage is calculated by the formula,

Molarity=Numberofmolesofsolute(n)Volumeofsolution(V)inL

Substitute the values of moles and volume for iron in the above formula of molarity as,

MolarityofFe=1.43molFe1L=1.43mol/L=1.43M_

Substitute the values of moles and volume for zinc in the above formula of molarity as,

MolarityofZn=9×102molZn1L=9.17×102mol/L=9.17×10-2M_

Therefore, the molarity of iron is 1.43M_ and the molarity of zinc in the drainage is 9.17×10-2M_ .

(b)

Interpretation Introduction

To determine: The complete chemical equation and a net ionic equation for the given process.

(b)

Expert Solution
Check Mark

Answer to Problem 4.132AP

Solution

The complete and balanced equation for the process is,

2Fe(OH)3(s)+3H2SO4(aq)Fe2(SO4)3(aq)+6H2O(l) .

The net ionic equation for the given process is,

2Fe(OH)3(s)+6H+(aq)2Fe3+(aq)+6H2O(l) .

Explanation of Solution

Explanation

The given in complete equation is,

2Fe(OH)3(s)+3H2SO4(aq)

The unbalanced equation is written as,

Fe(OH)3(s)+H2SO4(aq)Fe2(SO4)3(aq)+H2O(l) .

To balance the element iron in the above equation a coefficient of 2 is placed in front of Fe to the reactant side of the equation as,

2Fe(OH)3(s)+H2SO4(aq)Fe2(SO4)3(aq)+H2O(l) .

The oxygen and hydrogen atoms are balanced by placing coefficient of 3 in front of H2SO4 to the reactant side and a coefficient of 6 in front of water molecule to the product side of the equation as,

2Fe(OH)3(s)+3H2SO4(aq)Fe2(SO4)3(aq)+6H2O(l) .

The complete and balanced equation is,

2Fe(OH)3(s)+3H2SO4(aq)Fe2(SO4)3(aq)+6H2O(l) .

The net ionic equation of the above complete and balanced equation is obtained as follows.

Firstly, a balanced molecular equation is to be written that gives the overall ionic equation as,

2Fe(OH)3(s)+6H+(aq)+3SO42(aq)2Fe3+(aq)+3SO42(aq)+6H2O(l) .

The elimination of the spectator ions such as, 3SO42(aq) from the overall ionic equation will produce the net ionic equation as,

2Fe(OH)3(s)+6H+(aq)2Fe3+(aq)+6H2O(l) .

Therefore, the net ionic equation is,

2Fe(OH)3(s)+6H+(aq)2Fe3+(aq)+6H2O(l) .

(c)

Interpretation Introduction

To determine: The balanced net ionic equation for the reaction between smithsonite and H2SO4 that produces Zn2+(aq) .

(c)

Expert Solution
Check Mark

Answer to Problem 4.132AP

Solution

The balanced net ionic equation for the reaction between smithsonite and H2SO4 that produces Zn2+(aq) is,

ZnCO3(s)+2H+(aq)Zn2+(aq)+CO2(g)+H2O(l) .

Explanation of Solution

Explanation

The reaction between smithsonite (ZnCO3)   and H2SO4 produces zinc sulfate (ZnSO4(aq)) , carbon dioxide and water as,

ZnCO3(s)+H2SO4(aq)ZnSO4(aq)+CO2(g)+H2O(l)

The above equation is balanced. So there is no need to balance the atoms in the equation.

The net ionic equation of the above complete and balanced equation is obtained as follows.

Firstly, a balanced molecular equation is to be written that gives the overall ionic equation as,

ZnCO3(s)+2H+(aq)+SO42(aq)Zn2+(aq)+SO42(aq)+CO2(g)+H2O(l) .

The elimination of the spectator ions such as, SO42(aq) from the overall ionic equation will produce the net ionic equation as,

ZnCO3(s)+2H+(aq)Zn2+(aq)+CO2(g)+H2O(l) .

Therefore, the net ionic equation is,

ZnCO3(s)+2H+(aq)Zn2+(aq)+CO2(g)+H2O(l) .

(d)

Interpretation Introduction

To determine: The value of x in the given formula ZnxFe1x of the mineral in the deposit.

(d)

Expert Solution
Check Mark

Answer to Problem 4.132AP

Solution

The value of x in the given formula ZnxFe1x Fe2O3 of the mineral is 0.18_ .

Explanation of Solution

Explanation

The given formula is,

ZnxFe1x Fe2O3 .

By manipulating the above formula becomes as,

ZnxFe3xO4

The mole-mole conversion of the above formula is obtained by dividing moles of iron by mole of zinc as,

(3x)molFexmolZn (1)

The moles of iron (Fe) and zinc (Zn) are calculated by the formula,

Moles=Mass(g)Molarmass

Substitute the values of mass and molar mass of iron (Fe) in the above formula as,

Molesofiron(Fe)=80.0g55.85g×1molFe=1.43molFe

Substitute the values of mass and molar mass of zinc (Zn) in the above formula as,

Molesofzinc(Zn)=6.0g65.41g×1molZn=9.17×102molZn

Substitute the values of moles of iron and zinc in equation (1) as,

9.17×102molZn×(3x)molFexmolZn=1.43molFe=(9.17×102)(3x)=1.43x

The value of x is calculated as,

3(9.17×102)x(9.17×102)=1.43x(1.43+9.17×102)x=3(9.17×102)x=0.27511.5217x=0.18_

Therefore, the value of x in the given formula ZnxFe1x Fe2O3 of the mineral is 0.18_ .

Conclusion

  1. a. The molarity of iron is 1.43M_ and the molarity of zinc in the drainage is 9×10-2M_ .
  2. b. The complete and balanced equation for the process is,

    2Fe(OH)3(s)+3H2SO4(aq)Fe2(SO4)3(aq)+6H2O(l) .

    The net ionic equation for the given process is,

    2Fe(OH)3(s)+6H+(aq)2Fe3+(aq)+6H2O(l) .

  3. c. The balanced net ionic equation for the reaction between smithsonite and H2SO4 that produces Zn2+(aq) is,

    ZnCO3(s)+2H+(aq)Zn2+(aq)+CO2(g)+H2O(l) .

  4. d. The value of x in the given formula ZnxFe1x Fe2O3 of the mineral is 0.18_

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!

Chapter 4 Solutions

Chemistry

Ch. 4.7 - Prob. 11PECh. 4.7 - Prob. 12PECh. 4.7 - Prob. 13PECh. 4.7 - Prob. 14PECh. 4.7 - Prob. 15PECh. 4.9 - Prob. 16PECh. 4.9 - Prob. 17PECh. 4 - Prob. 4.1VPCh. 4 - Prob. 4.2VPCh. 4 - Prob. 4.3VPCh. 4 - Prob. 4.4VPCh. 4 - Prob. 4.5VPCh. 4 - Prob. 4.6VPCh. 4 - Prob. 4.7VPCh. 4 - Prob. 4.8VPCh. 4 - Prob. 4.9QPCh. 4 - Prob. 4.10QPCh. 4 - Prob. 4.11QPCh. 4 - Prob. 4.12QPCh. 4 - Prob. 4.13QPCh. 4 - Prob. 4.14QPCh. 4 - Prob. 4.15QPCh. 4 - Prob. 4.16QPCh. 4 - Prob. 4.17QPCh. 4 - Prob. 4.18QPCh. 4 - Prob. 4.19QPCh. 4 - Prob. 4.20QPCh. 4 - Prob. 4.21QPCh. 4 - Prob. 4.22QPCh. 4 - Prob. 4.23QPCh. 4 - Prob. 4.24QPCh. 4 - Prob. 4.25QPCh. 4 - Prob. 4.26QPCh. 4 - Prob. 4.27QPCh. 4 - Prob. 4.28QPCh. 4 - Prob. 4.29QPCh. 4 - Prob. 4.30QPCh. 4 - Prob. 4.31QPCh. 4 - Prob. 4.32QPCh. 4 - Prob. 4.33QPCh. 4 - Prob. 4.34QPCh. 4 - Prob. 4.35QPCh. 4 - Prob. 4.36QPCh. 4 - Prob. 4.37QPCh. 4 - Prob. 4.38QPCh. 4 - Prob. 4.39QPCh. 4 - Prob. 4.40QPCh. 4 - Prob. 4.41QPCh. 4 - Prob. 4.42QPCh. 4 - Prob. 4.43QPCh. 4 - Prob. 4.44QPCh. 4 - Prob. 4.45QPCh. 4 - Prob. 4.46QPCh. 4 - Prob. 4.47QPCh. 4 - Prob. 4.48QPCh. 4 - Prob. 4.49QPCh. 4 - Prob. 4.50QPCh. 4 - Prob. 4.51QPCh. 4 - Prob. 4.52QPCh. 4 - Prob. 4.53QPCh. 4 - Prob. 4.54QPCh. 4 - Prob. 4.55QPCh. 4 - Prob. 4.56QPCh. 4 - Prob. 4.57QPCh. 4 - Prob. 4.58QPCh. 4 - Prob. 4.59QPCh. 4 - Prob. 4.60QPCh. 4 - Prob. 4.61QPCh. 4 - Prob. 4.62QPCh. 4 - Prob. 4.63QPCh. 4 - Prob. 4.64QPCh. 4 - Prob. 4.65QPCh. 4 - Prob. 4.66QPCh. 4 - Prob. 4.67QPCh. 4 - Prob. 4.68QPCh. 4 - Prob. 4.69QPCh. 4 - Prob. 4.70QPCh. 4 - Prob. 4.71QPCh. 4 - Prob. 4.72QPCh. 4 - Prob. 4.73QPCh. 4 - Prob. 4.74QPCh. 4 - Prob. 4.75QPCh. 4 - Prob. 4.76QPCh. 4 - Prob. 4.77QPCh. 4 - Prob. 4.78QPCh. 4 - Prob. 4.79QPCh. 4 - Prob. 4.80QPCh. 4 - Prob. 4.81QPCh. 4 - Prob. 4.82QPCh. 4 - Prob. 4.83QPCh. 4 - Prob. 4.84QPCh. 4 - Prob. 4.85QPCh. 4 - Prob. 4.86QPCh. 4 - Prob. 4.87QPCh. 4 - Prob. 4.88QPCh. 4 - Prob. 4.89QPCh. 4 - Prob. 4.90QPCh. 4 - Prob. 4.91QPCh. 4 - Prob. 4.92QPCh. 4 - Prob. 4.93QPCh. 4 - Prob. 4.94QPCh. 4 - Prob. 4.95QPCh. 4 - Prob. 4.96QPCh. 4 - Prob. 4.97QPCh. 4 - Prob. 4.98QPCh. 4 - Prob. 4.99QPCh. 4 - Prob. 4.100QPCh. 4 - Prob. 4.101QPCh. 4 - Prob. 4.102QPCh. 4 - Prob. 4.103QPCh. 4 - Prob. 4.104QPCh. 4 - Prob. 4.105QPCh. 4 - Prob. 4.106QPCh. 4 - Prob. 4.107QPCh. 4 - Prob. 4.108QPCh. 4 - Prob. 4.109QPCh. 4 - Prob. 4.110QPCh. 4 - Prob. 4.111QPCh. 4 - Prob. 4.112QPCh. 4 - Prob. 4.113QPCh. 4 - Prob. 4.114QPCh. 4 - Prob. 4.115QPCh. 4 - Prob. 4.116QPCh. 4 - Prob. 4.117QPCh. 4 - Prob. 4.118QPCh. 4 - Prob. 4.119QPCh. 4 - Prob. 4.120QPCh. 4 - Prob. 4.122APCh. 4 - Prob. 4.123APCh. 4 - Prob. 4.124APCh. 4 - Prob. 4.125APCh. 4 - Prob. 4.126APCh. 4 - Prob. 4.127APCh. 4 - Prob. 4.128APCh. 4 - Prob. 4.129APCh. 4 - Prob. 4.130APCh. 4 - Prob. 4.131APCh. 4 - Prob. 4.132APCh. 4 - Prob. 4.133APCh. 4 - Prob. 4.134APCh. 4 - Prob. 4.135APCh. 4 - Prob. 4.136APCh. 4 - Prob. 4.137APCh. 4 - Prob. 4.138APCh. 4 - Prob. 4.139APCh. 4 - Prob. 4.140APCh. 4 - Prob. 4.141APCh. 4 - Prob. 4.142APCh. 4 - Prob. 4.143APCh. 4 - Prob. 4.144AP
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.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY