Chemistry: Atoms First
Chemistry: Atoms First
3rd Edition
ISBN: 9781259638138
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 26, Problem 26.56QP

(a)

Interpretation Introduction

Interpretation:

The balanced equations for the given aluminum compounds has to be described.

Concept introduction:

  • There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
  • The concept of writing a balanced chemical equation is depends on conversion of reactants into products.
  • First write the reaction from the given information.
  • Then count the number of atoms of each element in reactants as well as products.
  • Finally obtained values could place it as coefficients of reactants as well as products.

(a)

Expert Solution
Check Mark

Answer to Problem 26.56QP

The balanced equation for preparation of Al2Cl6 is 2Al(s) + 3Cl2(g) Al2Cl6(s)

Explanation of Solution

To write the balanced equation for preparation of Al2Cl6

We can produce Al2Cl6 by reaction of aluminum with chlorine gas.

Al(s) + Cl2(g) Al2Cl6(s)

Here, the reaction is unbalanced. So we need to balance it.  To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.  The balanced equation is

2Al(s) + 3Cl2(g) Al2Cl6(s)

Conclusion

The balanced equation for preparation of Al2Cl6 is 2Al(s) + 3Cl2(g) Al2Cl6(s)

(b)

Interpretation Introduction

Interpretation:

The balanced equations for the given aluminum compounds has to be described.

Concept introduction:

  • There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
  • The concept of writing a balanced chemical equation is depends on conversion of reactants into products.
  • First write the reaction from the given information.
  • Then count the number of atoms of each element in reactants as well as products.
  • Finally obtained values could place it as coefficients of reactants as well as products.

(b)

Expert Solution
Check Mark

Answer to Problem 26.56QP

The balanced equation for preparation of Al2O3 is 4Al(s) + 3O2(g) 2Al2O3(s)

Explanation of Solution

To write the balanced equation for preparation of Al2O3

We can produce Al2O3 by reaction of aluminum with oxygen gas.

Al(s) + O2(g) Al2O3(s)

Here, the reaction is unbalanced. So we need to balance it.  To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.  The balanced equation is

4Al(s) + 3O2(g) 2Al2O3(s)

Conclusion

The balanced equation for preparation of Al2O3 is 4Al(s) + 3O2(g) 2Al2O3(s)

(c)

Interpretation Introduction

Interpretation:

The balanced equations for the given aluminum compounds has to be described.

Concept introduction:

  • There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
  • The concept of writing a balanced chemical equation is depends on conversion of reactants into products.
  • First write the reaction from the given information.
  • Then count the number of atoms of each element in reactants as well as products.
  • Finally obtained values could place it as coefficients of reactants as well as products.

(c)

Expert Solution
Check Mark

Answer to Problem 26.56QP

The balanced equation for Al2(SO4)3 is 2Al(s) + 3H2SO4(aq)Al2(SO4)3(aq) + 3H2(g)

Explanation of Solution

To write the balanced equation for Al2(SO4)3

We can produce Al2(SO4)3 by reaction of aluminum with sulfuric acid.

Al(s) + H2SO4(aq)Al2(SO4)3(aq) + H2(g)

Here, the reaction is unbalanced. So we need to balance it.  To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.  The balanced equation is

2Al(s) + 3H2SO4(aq)Al2(SO4)3(aq) + 3H2(g)

Conclusion

The balanced equation for Al2(SO4)3 is 2Al(s) + 3H2SO4(aq)Al2(SO4)3(aq) + 3H2(g)

(d)

Interpretation Introduction

Interpretation:

The balanced equations for the given aluminum compounds has to be described.

Concept introduction:

  • There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
  • The concept of writing a balanced chemical equation is depends on conversion of reactants into products.
  • First write the reaction from the given information.
  • Then count the number of atoms of each element in reactants as well as products.
  • Finally obtained values could place it as coefficients of reactants as well as products.

(d)

Expert Solution
Check Mark

Answer to Problem 26.56QP

The balanced equation for NH4Al(SO4)2 is

Al2(SO4)3(aq) + (NH4)2SO4(aq)2NH4Al(SO4)2

Explanation of Solution

To write the balanced equation for NH4Al(SO4)2

We can produce NH4Al(SO4)2 by reaction of Al2(SO4)3(aq) with ammonium sulfate and after the evaporation process.

Al2(SO4)3(aq) + (NH4)2SO4(aq)NH4Al(SO4)2(s)

Here, the reaction is unbalanced. So we need to balance it.  To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.  The balanced equation is

Al2(SO4)3(aq) + (NH4)2SO4(aq)2NH4Al(SO4)2

Conclusion

The balanced equation for NH4Al(SO4)2 is

Al2(SO4)3(aq) + (NH4)2SO4(aq)2NH4Al(SO4)2

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Consider the series of reactions to synthesize the alum (KAl(SO4 )2 · xH2O(s)). ) Assuming an excess of the other reagents, from one mole of sulfuric acid H2SO4 , how many moles of alum will be produced?
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Chapter 26 Solutions

Chemistry: Atoms First

Ch. 26 - Prob. 26.11QPCh. 26 - Prob. 26.12QPCh. 26 - Prob. 26.13QPCh. 26 - Prob. 26.14QPCh. 26 - Prob. 26.15QPCh. 26 - Prob. 26.16QPCh. 26 - Prob. 26.17QPCh. 26 - Prob. 26.18QPCh. 26 - Which of the following compounds would require...Ch. 26 - Prob. 26.20QPCh. 26 - Prob. 26.21QPCh. 26 - Prob. 26.22QPCh. 26 - Prob. 26.23QPCh. 26 - Prob. 26.24QPCh. 26 - Prob. 26.25QPCh. 26 - Prob. 26.26QPCh. 26 - Prob. 26.27QPCh. 26 - Prob. 26.28QPCh. 26 - Prob. 26.29QPCh. 26 - Prob. 26.30QPCh. 26 - Prob. 26.31QPCh. 26 - Prob. 26.32QPCh. 26 - Prob. 26.33QPCh. 26 - Prob. 26.34QPCh. 26 - Prob. 26.35QPCh. 26 - Prob. 26.36QPCh. 26 - Prob. 26.37QPCh. 26 - Prob. 26.38QPCh. 26 - Prob. 26.39QPCh. 26 - Prob. 26.40QPCh. 26 - Prob. 26.41QPCh. 26 - Prob. 26.42QPCh. 26 - Prob. 26.43QPCh. 26 - Prob. 26.44QPCh. 26 - Prob. 26.45QPCh. 26 - Prob. 26.46QPCh. 26 - Prob. 26.47QPCh. 26 - With the Hall process, how many hours will it take...Ch. 26 - The overall reaction for the electrolytic...Ch. 26 - Prob. 26.50QPCh. 26 - Prob. 26.51QPCh. 26 - In basic solution, aluminum metal is a strong...Ch. 26 - Prob. 26.53QPCh. 26 - Prob. 26.54QPCh. 26 - Prob. 26.55QPCh. 26 - Prob. 26.56QPCh. 26 - Prob. 26.57QPCh. 26 - Prob. 26.58QPCh. 26 - Prob. 26.59QPCh. 26 - Prob. 26.60QPCh. 26 - Prob. 26.61QPCh. 26 - Prob. 26.62QPCh. 26 - Prob. 26.63QPCh. 26 - Prob. 26.64QPCh. 26 - Prob. 26.65QPCh. 26 - Prob. 26.66QPCh. 26 - Prob. 26.67QPCh. 26 - Prob. 26.68QPCh. 26 - Prob. 26.69QPCh. 26 - Prob. 26.70QPCh. 26 - Prob. 26.71QPCh. 26 - Prob. 26.72QPCh. 26 - Prob. 26.73QPCh. 26 - The following are two reaction schemes involving...Ch. 26 - Prob. 26.75QPCh. 26 - Prob. 26.76QPCh. 26 - Prob. 26.77QPCh. 26 - Prob. 26.78QPCh. 26 - Prob. 26.79QPCh. 26 - Prob. 26.80QPCh. 26 - Prob. 26.81QPCh. 26 - Chemical tests of four metals A, B, C, and D show...
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