EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781118930144
Author: Willard
Publisher: JOHN WILEY+SONS INC.
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Chapter 14, Problem 36PE

(a)

Interpretation Introduction

Interpretation:

Moles of Na2SO4 produced from 3.6 mol H2SO4 have to be determined.

Concept Introduction:

Stoichiometry describes quantitative relationships between reactants and products in any chemical reactions. In these problems, amount of one species is determined with help of known amount of another species.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given reaction occurs as follows:

  2NaOH(aq)+H2SO4(aq)Na2SO4(aq)+2H2O(l)

According to balanced chemical equation, two moles of NaOH react with one mole of H2SO4 and produce one mole of Na2SO4 and two moles of H2O. Since stoichiometric ratio between H2SO4 and Na2SO4 is 1:1, amount of Na2SO4 produced from 3.6 moles of H2SO4 is calculated as follows:

  Amount of Na2SO4=(1 mol Na2SO41 mol H2SO4)(3.6 mol H2SO4)=3.6 mol

Hence, amount of Na2SO4 is 3.6 mol.

(b)

Interpretation Introduction

Interpretation:

Moles of H2O produced from 0.025 mol NaOH have to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

Given reaction occurs as follows:

  2NaOH(aq)+H2SO4(aq)Na2SO4(aq)+2H2O(l)

According to balanced chemical equation, two moles of NaOH react with one mole of H2SO4 and produce one mole of Na2SO4 and two moles of H2O. Since stoichiometric ratio between NaOH and H2O is 2:2, amount of H2O produced from 0.025 moles of NaOH is calculated as follows:

  Amount of H2O=(2 mol H2O2 mol NaOH)(0.025 mol NaOH)=0.025 mol

Hence, amount of H2O is 0.025 mol.

(c)

Interpretation Introduction

Interpretation:

Moles of NaOH required to react with 2.50 L of 0.125 M H2SO4 have to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

Given reaction occurs as follows:

  2NaOH(aq)+H2SO4(aq)Na2SO4(aq)+2H2O(l)

Expression to calculate molarity of H2SO4 is as follows:

  Molarity of H2SO4=Moles of H2SO4Volume (L) of H2SO4 solution        (1)

Rearrange equation (1) for moles of H2SO4.

  Moles of H2SO4=[(Molarity of H2SO4)(Volume of H2SO4 solution)]        (2)

Substitute 0.125 M for molarity and 2.50 L for volume of H2SO4 solution in equation (2).

  Moles of H2SO4=(0.125 M)(2.50 L)=0.3125 mol

According to balanced chemical equation, two moles of NaOH react with one mole of H2SO4 and produce one mole of Na2SO4 and two moles of H2O. Since stoichiometric ratio between H2SO4 and NaOH is 1:2, amount of NaOH required to react with 0.3125 moles of H2SO4 is calculated as follows:

  Amount of NaOH=(2 mol NaOH1 mol H2SO4)(0.3125 mol H2SO4)=0.625 mol

Hence, amount of NaOH required is 0.625 mol.

(d)

Interpretation Introduction

Interpretation:

Grams of Na2SO4 obtained from 25 mL of 0.050 M NaOH have to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

Given reaction occurs as follows:

  2NaOH(aq)+H2SO4(aq)Na2SO4(aq)+2H2O(l)

Expression to calculate molarity of NaOH is as follows:

  Molarity of NaOH=Moles of NaOHVolume (L) of NaOH solution        (3)

Rearrange equation (3) for moles of NaOH.

  Moles of NaOH=[(Molarity of NaOH)(Volume of NaOH solution)]        (4)

Substitute 0.050 M for molarity and 25 mL for volume of NaOH solution in equation (4).

  Moles of NaOH=(0.050 M)(25 mL)(103 L1 mL)=0.00125 mol

According to balanced chemical equation, two moles of NaOH react with one mole of H2SO4 and produce one mole of Na2SO4 and two moles of H2O. Since stoichiometric ratio between Na2SO4 and NaOH is 1:2, amount of Na2SO4 required to react with 0.0625 moles of NaOH is calculated as follows:

  Amount of Na2SO4=(1 mol Na2SO42 mol NaOH)(0.00125 mol NaOH)=0.000625 mol

Expression for mass of Na2SO4 is as follows:

  Mass of Na2SO4=[(Moles of Na2SO4)(Molar mass of Na2SO4)]        (5)

Substitute 0.000625 mol for moles and 142.04 g/mol for molar mass of Na2SO4 in equation (5).

  Mass of Na2SO4=(0.000625 mol)(142.04 g/mol)=0.088775 g

Hence, 0.088775 g of Na2SO4 is required.

(e)

Interpretation Introduction

Interpretation:

Volume of 0.250 M H2SO4 required to react with 25.5 mL of 0.750 M NaOH have to be determined.

Concept Introduction:

Expression for molarity equation is as follows:

  M1V1=M2V2        (6)

Here,

M1 is molarity of one solution.

V1 is volume of one solution.

M2 is molarity of another solution.

V2 is volume of another solution.

(e)

Expert Solution
Check Mark

Explanation of Solution

Rearrange equation (6) for V2.

  V2=M1V1M2        (7)

Substitute 0.750 M for M1, 25.5 mL for V1 and 0.250 M for M2 in equation (7) for volume of H2SO4.

  Volume of H2SO4=(0.750 M)(25.5 mL)0.250 M=76.5 mL

Hence volume of H2SO4 required is 76.5 mL.

(f)

Interpretation Introduction

Interpretation:

Molarity of NaOH formed by reaction of 48.20 mL of NaOH with 35.72 mL of 0.125 M H2SO4 have to be determined.

Concept Introduction:

Refer to part (e).

(f)

Expert Solution
Check Mark

Explanation of Solution

Rearrange equation (6) for M2.

  M2=M1V1V2        (8)

Substitute 0.125 M for M1, 48.20 mL for V1 and 35.72 mL for V2 in equation (8) for molarity of NaOH.

  Molarity of NaOH=(0.125 M)(48.20 mL)35.72 mL=0.169 M

Hence molarity of NaOH required is 0.169 M.

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Chapter 14 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

Ch. 14.5 - Prob. 14.11PCh. 14.5 - Prob. 14.12PCh. 14 - Prob. 1RQCh. 14 - Prob. 2RQCh. 14 - Prob. 3RQCh. 14 - Prob. 4RQCh. 14 - Prob. 5RQCh. 14 - Prob. 6RQCh. 14 - Prob. 7RQCh. 14 - Prob. 8RQCh. 14 - Prob. 9RQCh. 14 - Prob. 10RQCh. 14 - Prob. 11RQCh. 14 - Prob. 12RQCh. 14 - Prob. 13RQCh. 14 - Prob. 14RQCh. 14 - Prob. 15RQCh. 14 - Prob. 16RQCh. 14 - Prob. 17RQCh. 14 - Prob. 18RQCh. 14 - Prob. 19RQCh. 14 - Prob. 20RQCh. 14 - Prob. 21RQCh. 14 - Prob. 22RQCh. 14 - Prob. 23RQCh. 14 - Prob. 24RQCh. 14 - Prob. 25RQCh. 14 - Prob. 26RQCh. 14 - Prob. 27RQCh. 14 - Prob. 28RQCh. 14 - Prob. 29RQCh. 14 - Prob. 30RQCh. 14 - Prob. 31RQCh. 14 - Prob. 32RQCh. 14 - Prob. 33RQCh. 14 - Prob. 34RQCh. 14 - Prob. 35RQCh. 14 - Prob. 37RQCh. 14 - Prob. 38RQCh. 14 - Prob. 39RQCh. 14 - Prob. 40RQCh. 14 - Prob. 41RQCh. 14 - Prob. 42RQCh. 14 - Prob. 1PECh. 14 - Prob. 2PECh. 14 - Prob. 3PECh. 14 - Prob. 4PECh. 14 - Prob. 5PECh. 14 - Prob. 6PECh. 14 - Prob. 7PECh. 14 - Prob. 8PECh. 14 - Prob. 9PECh. 14 - Prob. 10PECh. 14 - Prob. 11PECh. 14 - Prob. 12PECh. 14 - Prob. 13PECh. 14 - Prob. 14PECh. 14 - Prob. 15PECh. 14 - Prob. 16PECh. 14 - Prob. 17PECh. 14 - Prob. 18PECh. 14 - Prob. 19PECh. 14 - Prob. 20PECh. 14 - Prob. 21PECh. 14 - Prob. 22PECh. 14 - Prob. 23PECh. 14 - Prob. 24PECh. 14 - Prob. 25PECh. 14 - Prob. 26PECh. 14 - Prob. 27PECh. 14 - Prob. 28PECh. 14 - Prob. 29PECh. 14 - Prob. 30PECh. 14 - Prob. 31PECh. 14 - Prob. 32PECh. 14 - Prob. 33PECh. 14 - Prob. 34PECh. 14 - Prob. 35PECh. 14 - Prob. 36PECh. 14 - Prob. 37PECh. 14 - Prob. 38PECh. 14 - Prob. 39PECh. 14 - Prob. 40PECh. 14 - Prob. 41PECh. 14 - Prob. 42PECh. 14 - Prob. 44PECh. 14 - Prob. 45PECh. 14 - Prob. 46PECh. 14 - Prob. 47PECh. 14 - Prob. 48PECh. 14 - Prob. 49PECh. 14 - Prob. 50PECh. 14 - Prob. 51PECh. 14 - Prob. 52PECh. 14 - Prob. 53AECh. 14 - Prob. 54AECh. 14 - Prob. 55AECh. 14 - Prob. 56AECh. 14 - Prob. 57AECh. 14 - Prob. 58AECh. 14 - Prob. 59AECh. 14 - Prob. 60AECh. 14 - Prob. 61AECh. 14 - Prob. 62AECh. 14 - Prob. 63AECh. 14 - Prob. 65AECh. 14 - Prob. 66AECh. 14 - Prob. 67AECh. 14 - Prob. 68AECh. 14 - Prob. 69AECh. 14 - Prob. 70AECh. 14 - Prob. 71AECh. 14 - Prob. 72AECh. 14 - Prob. 73AECh. 14 - Prob. 74AECh. 14 - Prob. 75AECh. 14 - Prob. 76AECh. 14 - Prob. 77AECh. 14 - Prob. 78AECh. 14 - Prob. 79AECh. 14 - Prob. 80AECh. 14 - Prob. 81AECh. 14 - Prob. 82AECh. 14 - Prob. 83AECh. 14 - Prob. 84AECh. 14 - Prob. 85AECh. 14 - Prob. 86AECh. 14 - Prob. 87AECh. 14 - Prob. 88AECh. 14 - Prob. 90AECh. 14 - Prob. 91AECh. 14 - Prob. 92AECh. 14 - Prob. 93AECh. 14 - Prob. 94AECh. 14 - Prob. 95AECh. 14 - Prob. 96AECh. 14 - Prob. 97AECh. 14 - Prob. 98AECh. 14 - Prob. 99CECh. 14 - Prob. 100CECh. 14 - Prob. 102CECh. 14 - Prob. 103CECh. 14 - Prob. 104CECh. 14 - Prob. 105CE
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