EBK INTRODUCTION TO CHEMISTRY
EBK INTRODUCTION TO CHEMISTRY
5th Edition
ISBN: 9781260162165
Author: BAUER
Publisher: MCGRAW HILL BOOK COMPANY
bartleby

Videos

Question
Book Icon
Chapter 11, Problem 115QP

(a)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given Information:

The molarity of barium chloride solution is 0.100 M and of potassium sulfate solution is 0.500 M . The volume of potassium sulfate solution is 90 ml and of barium chloride solution is 500 ml .

Titration is a method to determine the concentration of a substance in the solution by making it react with a solution of known concentration of other substance, just beyond the point where the reaction between both the substances completes. In precipitation reactions, on the reaction of the reactants, an insoluble end product is formed which precipitates out from the solution.

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

BaCl2aq+K2SO4aqBaSO4s+2KClaq

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

1 L=1000 mL1 mL=1 L1000 mL

Convert 90 ml volume of potassium sulfate solution from milliliters to liters as follows:

90 mL=1 L1000 mL×90 mL=0.09L

Convert 500 ml volume of barium chloride solution from milliliters to liters as follows:

500 mL=1 L1000 mL×500 mL=0.5L

Molarity of the solution M is equal to the ratio of the number of moles n of a solute to the volume V of the solution in litres.

M=nV …… (1)

Substitute M as 0.1 mol L1 and V as 0.5 L in equation (1) to determine the number of moles of barium chloride as follows:

0.1 mol L1=n0.5Ln=0.1 mol L1×0.5L=0.05 moles

Substitute M as 0.5 mol L1 and V as 0.09 L in equation (1) to determine the number of moles of potassium sulphate as follows:

0.5 mol L1=n0.09 Ln=0.5 mol L1×0.09L=0.045 moles

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate. Therefore,

1 mol BaCl2 reacts with=1 mol K2SO40.05 mol BaCl2 reacts with=1 mol K2SO41 mol BaCl2×0.05 mol BaCl2=0.05 mol K2SO4

Barium sulfate formed is as follows:

1 mol K2SO4 forms=1 mol BaSO40.05 mol K2SO4 forms=1 mol BaSO41 mol K2SO4×0.05 mol K2SO4=0.05 mol BaSO4

However, only 0.045 mol of K2SO4 is present. Therefore, 0.045 mol of K2SO4 reacts with 0.045 mol of BaCl2 to produce 0.045 mol of BaSO4 .

The molar mass of BaSO4 is as follows:

BaSO4=137.38 g mol1+32 g mol1+4×16 g mol1=233.38 g mol1

The number of moles n of a substance is equal to the ratio of the mass m of that substance to the molar mass MM of the substance.

n=mMM …… (2)

Substitute n as 0.045 moles and MM as 233.38 g mol1 in equation (2) to determine the mass of BaSO4 as follows:

0.045 mol=m233.38 g mol1m=0.045 mol×233.38g mol1=10.5021 g

Thus, the mass of barium sulfate formed is l0.502l g .

(b)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given Information:

The molarity of barium chloride solution is 0.100 M and of potassium sulfate solution is 0.500 M . The volume of potassium sulfate solution is 100 ml and of barium chloride solution is 100 ml .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

BaCl2aq+K2SO4aqBaSO4s+2KClaq

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

1 L=1000 mL1 mL=1 L1000 mL

Convert 100 ml volume of potassium sulfate solution from milliliters to liters as follows:

100 mL=1 L1000 mL×100 mL=0.01L

Convert 100 ml volume of barium chloride solution from milliliters to liters as follows:

100 mL=1 L1000 mL×100 mL=0.1L

Substitute M as 0.1 mol L1 and V as 0.1 L in equation (1) to determine the number of moles of barium chloride as follows:

0.1 mol L1=n0.1Ln=0.1 mol L1×0.1 L=0.01 moles

Substitute M as 0.5 mol L1 and V as 0.1 L in equation (1) to determine the number of moles of potassium sulfate as follows:

0.5 mol L1=n0.1Ln=0.5 mol L1×0.1 L=0.05 moles

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

1 mol BaCl2 reacts=1 mol K2SO40.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4

The number of moles of BaSO4 produced as follows:

1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4

The molar mass of BaSO4 is as follows:

BaSO4=137.38 g mol1+32 g mol1+4×16 g mol1=233.38 g mol1

Substitute n as 0.01 moles and MM as 233.38 g mol1 in equation (2) to determine the mass of BaSO4 as follows:

0.01 mol=m233.38 g mol1m=0.01 mol×233.38g mol1=2.3338 g

Thus, the mass of barium sulfate formed is 2.3338 g .

(c)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given Information:

The molarity of barium chloride solution is 0.100 M and of potassium sulfate solution is 0.500 M . The volume of potassium sulfate solution is 500 ml and of barium chloride solution is 100 ml .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

BaCl2aq+K2SO4aqBaSO4s+2KClaq

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

1 L=1000 mL1 mL=1 L1000 mL

Convert 500 ml volume of potassium sulfate solution from milliliters to liters as follows:

500 mL=1 L1000 mL×500 mL=0.5L

Convert 100 ml volume of barium chloride solution from milliliters to liters as follows:

100 mL=1 L1000 mL×100 mL=0.1L

Substitute M as 0.1 mol L1 and V as 0.1 L in equation (1) to determine the number of moles of barium chloride as follows:

0.1 mol L1=n0.1Ln=0.1 mol L1×0.1 L=0.01 moles

Substitute M as 0.5 mol L1 and V as 0.5 L in equation (1) to determine the number of moles of potassium sulfate as follows:

0.5 mol L1=n0.5Ln=0.5 mol L1×0.5 L=0.25 moles

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

 0.25 moles of K2SO4 must react with 0.25 mole of BaCl2 to form 0.25 mole of BaSO4But there are only 0.01 mole of BaCl2 0.01 moles of BaCl2  reacts with 0.01 moles of K2 SO4  to form 0.01 moles of BaSO4

1 mol BaCl2 reacts=1 mol K2SO40.25 mol K2SO4 reacts=1 mol BaCl21 mol K2SO4×0.25 mol K2SO4=0.25 mol BaCl2

However, the number of moles of BaCl2 is only 0.01 mol .

0.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4

The amount of BaSO4 formed is as follows:

1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4

The molar mass of BaSO4 is as follows:

BaSO4=137.38 g mol1+32 g mol1+4×16 g mol1=233.38 g mol1

Substitute n as 0.01 moles and MM as 233.38 g mol1 in equation (2) to determine the mass of BaSO4 as follows:

0.01 mol=m233.38 g mol1m=0.01 mol×233.38g mol1=2.3338 g

Thus, the mass of barium sulfate formed is 2.3338 g .

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
(f) SO: Best Lewis Structure 3 e group geometry:_ shape/molecular geometry:, (g) CF2CF2 Best Lewis Structure polarity: e group arrangement:_ shape/molecular geometry: (h) (NH4)2SO4 Best Lewis Structure polarity: e group arrangement: shape/molecular geometry: polarity: Sketch (with angles): Sketch (with angles): Sketch (with angles):
1. Problem Set 3b Chem 141 For each of the following compounds draw the BEST Lewis Structure then sketch the molecule (showing bond angles). Identify (i) electron group geometry (ii) shape around EACH central atom (iii) whether the molecule is polar or non-polar (iv) (a) SeF4 Best Lewis Structure e group arrangement:_ shape/molecular geometry: polarity: (b) AsOBr3 Best Lewis Structure e group arrangement:_ shape/molecular geometry: polarity: Sketch (with angles): Sketch (with angles):
(c) SOCI Best Lewis Structure 2 e group arrangement: shape/molecular geometry:_ (d) PCls Best Lewis Structure polarity: e group geometry:_ shape/molecular geometry:_ (e) Ba(BrO2): Best Lewis Structure polarity: e group arrangement: shape/molecular geometry: polarity: Sketch (with angles): Sketch (with angles): Sketch (with angles):

Chapter 11 Solutions

EBK INTRODUCTION TO CHEMISTRY

Ch. 11 - Prob. 5PPCh. 11 - Prob. 6PPCh. 11 - Prob. 7PPCh. 11 - Prob. 8PPCh. 11 - Prob. 9PPCh. 11 - Prob. 10PPCh. 11 - Prob. 11PPCh. 11 - Prob. 1QPCh. 11 - Prob. 2QPCh. 11 - Prob. 3QPCh. 11 - Prob. 4QPCh. 11 - Prob. 5QPCh. 11 - Prob. 6QPCh. 11 - Prob. 7QPCh. 11 - Prob. 8QPCh. 11 - Prob. 9QPCh. 11 - Prob. 10QPCh. 11 - Prob. 11QPCh. 11 - Prob. 12QPCh. 11 - Prob. 13QPCh. 11 - Prob. 14QPCh. 11 - Prob. 15QPCh. 11 - Prob. 16QPCh. 11 - Prob. 17QPCh. 11 - Prob. 18QPCh. 11 - Prob. 19QPCh. 11 - When NaOH dissolves in water, the solution feels...Ch. 11 - Prob. 21QPCh. 11 - Prob. 22QPCh. 11 - Prob. 23QPCh. 11 - Prob. 24QPCh. 11 - Prob. 25QPCh. 11 - Prob. 26QPCh. 11 - Use the rue “like dissolves like� to predict...Ch. 11 - Prob. 28QPCh. 11 - Prob. 29QPCh. 11 - Prob. 30QPCh. 11 - Use intermolecular forces to explain why NaCl is...Ch. 11 - Prob. 32QPCh. 11 - Prob. 33QPCh. 11 - Prob. 34QPCh. 11 - Prob. 35QPCh. 11 - Prob. 36QPCh. 11 - Prob. 37QPCh. 11 - Prob. 38QPCh. 11 - Prob. 39QPCh. 11 - Prob. 40QPCh. 11 - Prob. 41QPCh. 11 - Prob. 42QPCh. 11 - Prob. 43QPCh. 11 - Prob. 44QPCh. 11 - How might you prepare a saturated solution of a...Ch. 11 - Prob. 46QPCh. 11 - Prob. 47QPCh. 11 - Prob. 48QPCh. 11 - Prob. 49QPCh. 11 - Prob. 50QPCh. 11 - Prob. 51QPCh. 11 - Prob. 52QPCh. 11 - Prob. 53QPCh. 11 - Prob. 54QPCh. 11 - Prob. 55QPCh. 11 - Prob. 56QPCh. 11 - Prob. 57QPCh. 11 - Prob. 58QPCh. 11 - The chemical trichloroethylene (TCE) is a...Ch. 11 - Prob. 60QPCh. 11 - Prob. 61QPCh. 11 - Prob. 62QPCh. 11 - Prob. 63QPCh. 11 - Prob. 64QPCh. 11 - Prob. 65QPCh. 11 - Prob. 66QPCh. 11 - Prob. 67QPCh. 11 - Prob. 68QPCh. 11 - Drinking water may contain a low concentration of...Ch. 11 - Prob. 70QPCh. 11 - Prob. 71QPCh. 11 - Prob. 72QPCh. 11 - Prob. 73QPCh. 11 - Prob. 74QPCh. 11 - Prob. 75QPCh. 11 - Prob. 76QPCh. 11 - Prob. 77QPCh. 11 - Prob. 78QPCh. 11 - Prob. 79QPCh. 11 - Prob. 80QPCh. 11 - Prob. 81QPCh. 11 - Prob. 82QPCh. 11 - Prob. 83QPCh. 11 - Prob. 84QPCh. 11 - Prob. 85QPCh. 11 - Prob. 86QPCh. 11 - Prob. 87QPCh. 11 - Prob. 88QPCh. 11 - Prob. 89QPCh. 11 - Prob. 90QPCh. 11 - Prob. 91QPCh. 11 - Prob. 92QPCh. 11 - Prob. 93QPCh. 11 - Prob. 94QPCh. 11 - Prob. 95QPCh. 11 - Prob. 96QPCh. 11 - Prob. 97QPCh. 11 - Prob. 98QPCh. 11 - Prob. 99QPCh. 11 - Prob. 100QPCh. 11 - Prob. 101QPCh. 11 - Prob. 102QPCh. 11 - Prob. 103QPCh. 11 - Prob. 104QPCh. 11 - Prob. 105QPCh. 11 - Prob. 106QPCh. 11 - Prob. 107QPCh. 11 - The solubility of KNO3 increases as the...Ch. 11 - Prob. 109QPCh. 11 - Prob. 110QPCh. 11 - Prob. 111QPCh. 11 - Prob. 112QPCh. 11 - Prob. 113QPCh. 11 - Prob. 114QPCh. 11 - Prob. 115QPCh. 11 - Prob. 116QPCh. 11 - Prob. 117QPCh. 11 - Prob. 118QPCh. 11 - Prob. 119QPCh. 11 - Prob. 120QPCh. 11 - A salad dressing can be made by shaking together...Ch. 11 - Prob. 122QPCh. 11 - Prob. 123QPCh. 11 - Prob. 124QPCh. 11 - Prob. 125QPCh. 11 - Prob. 126QPCh. 11 - Prob. 127QPCh. 11 - Prob. 128QPCh. 11 - Prob. 129QPCh. 11 - Prob. 130QPCh. 11 - Prob. 131QPCh. 11 - Prob. 132QPCh. 11 - Prob. 133QPCh. 11 - Lead(II) iodide, PbI2, is a yellow solid with a...Ch. 11 - Prob. 135QPCh. 11 - Prob. 136QPCh. 11 - Prob. 137QPCh. 11 - Prob. 138QPCh. 11 - Prob. 139QP
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
    General Chemistry - Standalone book (MindTap Cour...
    Chemistry
    ISBN:9781305580343
    Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
    Publisher:Cengage Learning
    Text book image
    Chemistry: The Molecular Science
    Chemistry
    ISBN:9781285199047
    Author:John W. Moore, Conrad L. Stanitski
    Publisher:Cengage Learning
    Text book image
    Chemistry: Principles and Practice
    Chemistry
    ISBN:9780534420123
    Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
    Publisher:Cengage Learning
  • Text book image
    Chemistry
    Chemistry
    ISBN:9781305957404
    Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    Text book image
    Chemistry: An Atoms First Approach
    Chemistry
    ISBN:9781305079243
    Author:Steven S. Zumdahl, Susan A. Zumdahl
    Publisher:Cengage Learning
    Text book image
    Chemistry
    Chemistry
    ISBN:9781133611097
    Author:Steven S. Zumdahl
    Publisher:Cengage Learning
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Thermogravimetric Analysis [ TGA ] # Thermal Analysis # Analytical Chemistry Part-11# CSIR NET/GATE; Author: Priyanka Jain;https://www.youtube.com/watch?v=p1K-Jpzylso;License: Standard YouTube License, CC-BY