Principles of General Chemistry
Principles of General Chemistry
3rd Edition
ISBN: 9780073402697
Author: SILBERBERG, Martin S.
Publisher: McGraw-Hill College
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Chapter 4, Problem 4.33P

a)

Interpretation Introduction

Interpretation:Balanced total and net equations for reaction of aluminum sulfate with aqueous NaOH should be determined.

Concept introduction:Three types of equations are used for representation of ionic reactions. These are mentioned below.

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

Molecular equations indicate reactants and products in undissociated forms. In total ionic equations, total ions that are dissociated in reaction are represented whereas in net ionic reactions, only useful ions are written.

Chemical reactions that lead to formation of insoluble salt by combination of two soluble salts with each other are called precipitation reactions. Such an insoluble salt is called precipitate. These reactions are so called due to formation of precipitates in them.

a)

Expert Solution
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Explanation of Solution

In this reaction, Al2( SO4)3 reacts with NaOH and forms Na3SO4 and Al(OH)3 . Since common sulfates are soluble but hydroxides are insoluble, precipitate of Al(OH)3 isformed in this reaction. So this reaction is example of precipitation reaction. Chemcal equation for this as follows:

  Al2( SO4)3(aq)+6NaOH(aq)3Na2SO4(aq)+2Al(OH)3(s)

Species present in aqueous phases are dissociated into their respective ions to write total ionic equation as follows:

  2Al3+(aq)+3SO42(aq)+6Na+(aq)+6OH(aq)(6Na+( aq)+3SO4 2( aq)+2Al( OH)3(s))

  Na+ and SO42 are common ions present in the reaction mixture. Therefore net ionic equation for given reaction becomes,

  2Al3+(aq)+6OH(aq)2Al(OH)3(s)

b)

Interpretation Introduction

Interpretation:Mass of precipitate formed by addition of 185.5 mL of 0.533 MNaOH to 627 mL of solution that contains 15.8 g of aluminum sulfate per liter is to be determined.

Concept introduction:Quantitative relationships between reactants and products used in chemical reactions are determined by stoichiometry. In such problems, amount of one species is given and that of another species is to be calculated.

Reagent that is fully consumed in any chemical reaction is known as limiting reagent. Yield of product is decided by its presence only. Other species present in any chemical reaction other than limiting reactant are present in excess amount and remain unreacted at end of reaction.

b)

Expert Solution
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Answer to Problem 4.33P

Mass of Al(OH)3 formed by addition of 185.5 mL of 0.533 MNaOH is added to 627 mL of a solution that contains 15.8 g of aluminum sulfate per liter is 2.57 g .

Explanation of Solution

Chemical equation for given reaction is as follows:

  Al2( SO4)3(aq)+6NaOH(aq)3Na2SO4(aq)+2Al(OH)3(s)

Formula to calculate moles NaOH of is as follows:

  MolesofNaOH=(Molarity of NaOH)(Volume ofNaOH)

Molarity of NaOH is 0.533M .

Volume of NaOH is 185.5mL .

Substitute the values in above equation.

  MolesofNaOH=(Molarity of NaOH)(Volume ofNaOH)=(0.533M)(185.5 mL)( 10 3  L 1 mL)=0.0988715 mol

Since six moles of NaOH forms two moles of Al(OH)3 , moles of Al(OH)3 formed in this reaction are calculated as follows:

  MolesofAl(OH)3=(0.0988715 mol NaOH)( 2molAl ( OH ) 3 6molNaOH)=0.03296 mol

Formula to calculate mass of Al(OH)3 is as follows:

  Mass of Al(OH)3=(Moles of Al( OH)3)(Molar mass of Al( OH)3)

Moles of Al(OH)3 are 0.03296 mol .

Molar mass of Al(OH)3 is 78.00 g/mol .

Substitute the values in above equation.

  Mass of Al(OH)3=(Moles of Al ( OH )3)(Molar mass of Al ( OH )3)=(0.03296 mol)(78.00 g/mol)=2.57088 g2.57 g

Moles of Al2( SO4)3 are calculated as follows:

  Moles of Al2( SO 4)3=( 15.8 g 342.14 g/mL)(627 mL)( 10 3  L 1 mL)=0.02895 mol

Since one mole of Al2( SO4)3 forms two moles of Al(OH)3 , moles of Al(OH)3 formed in this reaction are calculated as follows:

  MolesofAl(OH)3=(0.02895  mol Al2 ( SO 4 )3)( 2molAl ( OH ) 3 1mol Al 2 ( SO 4 ) 3 )=0.0579 mol

Formula to calculate mass of Al(OH)3 is as follows:

  Mass of Al(OH)3=(Moles of Al( OH)3)(Molar mass of Al( OH)3)

Moles of Al(OH)3 are 0.0579 mol .

Molar mass of Al(OH)3 is 78.00 g/mol .

Substitute the values in above equation.

  Mass of Al(OH)3=(Moles of Al ( OH )3)(Molar mass of Al ( OH )3)=(0.0579 mol)(78.00 g/mol)=4.5162 g4.52 g

Since mass of Al(OH)3 formed by NaOH is less than that formed by Al2( SO4)3 , NaOH acts as limiting reactant. Therefore mass of Al(OH)3 formed is 2.57 g .

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Chapter 4 Solutions

Principles of General Chemistry

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