Chapter 9, Problem 8A

(a)

Interpretation Introduction

Interpretation: The number of moles of products formed during the reaction of 0.125 moles of $\text{FeO}$ in the given reaction needs to be determined.

  ${\text{FeO}}_{\text{(s) }}{\text{+ C}}_{\text{(s)}}\stackrel{}{\to }{\text{ Fe}}_{\text{(l)}}{\text{ + CO}}_{\text{2(g)}}$

Concept Introduction: A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant.

Answer to Problem 8A

  $\text{0}{\text{.125 moles Fe}}_{\text{(l)}}$ and $\text{0}{\text{.0625 moles CO}}_{\text{2}}{}_{\text{(g)}}$

Explanation of Solution

The balance chemical equation:

  ${\text{2 FeO}}_{\text{(s) }}{\text{+ C}}_{\text{(s)}}\stackrel{}{\to }{\text{ 2 Fe}}_{\text{(l)}}{\text{ + CO}}_{\text{2(g)}}$

Moles of $\text{FeO}$ = 0.125moles

From the balance chemical equation:

2 mole of $\text{FeO}$ = form 2 mole of $\text{Fe}$ = 1 mole of ${\text{CO}}_{\text{2}}$

Calculate moles of $\text{Fe}$ and ${\text{CO}}_{\text{2}}$ :

  $\begin{array}{l}\text{0}{\text{.125 moles FeO}}_{\text{(s) }}\text{×}\frac{{\text{2 mole Fe}}_{\text{(l)}}}{\text{2 mole FeO}}\text{ = 0}{\text{.125 moles Fe}}_{\text{(l)}}\\ \text{0}{\text{.125 moles FeO}}_{\text{(s) }}\text{×}\frac{{\text{1 mole CO}}_{\text{2}}{}_{\text{(g)}}}{\text{2 mole FeO}}\text{ = 0}{\text{.0625 moles CO}}_{\text{2}}{}_{\text{(g)}}\end{array}$

(b)

Interpretation Introduction

Interpretation: The number of moles of products formed during the reaction of 0.125 moles of $\text{KI}$ in the given reaction needs to be determined.

  ${\text{Cl}}_{\text{2}}{}_{\text{(g) }}{\text{+ KI}}_{\text{(aq)}}\stackrel{}{\to }{\text{ KCl}}_{\text{(aq)}}{\text{ + I}}_{\text{2(s)}}$

Concept Introduction: A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant.

Answer to Problem 8A

  $\text{ 0}{\text{.125 moles KI}}_{\text{(aq) }}$ and $\text{0}{\text{.0625 moles I}}_{\text{2}}{}_{\text{(s)}}$

Explanation of Solution

The balance chemical equation:

  ${\text{Cl}}_{\text{2}}{}_{\text{(g) }}{\text{+2 KI}}_{\text{(aq)}}\stackrel{}{\to }{\text{ 2 KCl}}_{\text{(aq)}}{\text{ + I}}_{\text{2(s)}}$

Moles of $\text{KI}$ = 0.125 moles

From the balance chemical equation:

2 mole of $\text{KI}$ = form 2 mole of $\text{KCl}$ = 1 mole of ${\text{I}}_{\text{2}}$

Calculate moles of $\text{KCl}$ and ${\text{I}}_{\text{2}}$ :

  $\begin{array}{l}\text{0}{\text{.125 moles KI}}_{\text{(aq) }}\text{×}\frac{{\text{2 mole KCl}}_{\text{(aq)}}}{{\text{2 mole KI}}_{\text{(aq) }}}\text{ = 0}{\text{.125 moles KI}}_{\text{(aq) }}\\ \text{0}{\text{.125 moles KI}}_{\text{(aq) }}\text{×}\frac{{\text{1 mole I}}_{\text{2}}{}_{\text{(s)}}}{{\text{2 mole KI}}_{\text{(aq) }}}\text{ = 0}{\text{.0625 moles I}}_{\text{2}}{}_{\text{(s)}}\end{array}$

(c)

Interpretation Introduction

Interpretation: The moles of products formed during the reaction of 0.125 moles of ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}$ in the given reaction needs to be determined.

  ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}{}_{\text{(s) }}{\text{+ H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}{\text{+ 5 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{ 4 H}}_{\text{3}}{\text{BO}}_{\text{3}}{}_{\text{(s)}}{\text{ + Na}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}$

Concept Introduction: A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant.

Answer to Problem 8A

  $\text{0}{\text{.50 moles H}}_{\text{3}}{\text{BO}}_{\text{3}}{}_{\text{(s)}}$ and $\text{ 0}{\text{.125 moles Na}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}$

Explanation of Solution

The balance chemical equation:

  ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}{}_{\text{(s) }}{\text{+ H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}{\text{+ 5 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{ 4 H}}_{\text{3}}{\text{BO}}_{\text{3}}{}_{\text{(s)}}{\text{ + Na}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}$

Moles of ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}$ = 0.125 moles

From the balance chemical equation:

1 mole of ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}$ = form 4 mole of ${\text{ H}}_{\text{3}}{\text{BO}}_{\text{3}}{}_{\text{(s)}}$ = 1 mole of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}$

Calculate moles of ${\text{ H}}_{\text{3}}{\text{BO}}_{\text{3}}{}_{\text{(s)}}$ and ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}$ :

  $\begin{array}{l}\text{0}{\text{.125 moles Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\text{×}\frac{{\text{4 mole H}}_{\text{3}}{\text{BO}}_{\text{3}}{}_{\text{(s)}}}{{\text{1 mole Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}}\text{ = 0}{\text{.50 moles H}}_{\text{3}}{\text{BO}}_{\text{3}}{}_{\text{(s)}}\\ \text{0}{\text{.125 moles Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\text{×}\frac{{\text{1 mole Na}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}}{{\text{1 mole Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}}\text{ = 0}{\text{.125 moles Na}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}\end{array}$

(d)

Interpretation Introduction

Interpretation: The moles of products formed during the reaction of 0.125 moles of ${\text{CaC}}_{\text{2}}$ in the given reaction needs to be determined.

  ${\text{CaC}}_{\text{2}}{}_{\text{(s) }}{\text{+ 2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{ Ca(OH)}}_{\text{2}}{}_{\text{(s)}}{\text{ + C}}_{\text{2}}{\text{H}}_2{}_{\text{(g)}}$

Concept Introduction: A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant.

Answer to Problem 8A

  $\text{0}{\text{.125 moles Ca(OH)}}_{\text{2}}{}_{\text{(s)}}$ and $\text{0}{\text{.125 moles C}}_{\text{2}}{\text{H}}_{\text{2}}{}_{\text{(g)}}$

Explanation of Solution

The balance chemical equation:

  ${\text{CaC}}_{\text{2}}{}_{\text{(s) }}{\text{+ 2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{ Ca(OH)}}_{\text{2}}{}_{\text{(s)}}{\text{ + C}}_{\text{2}}{\text{H}}_2{}_{\text{(g)}}$

Moles of ${\text{CaC}}_{\text{2}}$ = 0.125 moles

From the balance chemical equation:

2 mole of ${\text{CaC}}_{\text{2}}$ = form 1 mole of ${\text{Ca(OH)}}_{\text{2}}{}_{\text{(s)}}$ = 1 mole of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\text{(g)}}$

Calculate moles of ${\text{Ca(OH)}}_{\text{2}}{}_{\text{(s)}}$ and ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\text{(g)}}$ :

  $\begin{array}{l}\text{0}{\text{.125 moles CaC}}_{\text{2}}\text{×}\frac{{\text{1 mole Ca(OH)}}_{\text{2}}{}_{\text{(s)}}}{{\text{1 mole CaC}}_{\text{2}}}\text{ = 0}{\text{.125 moles Ca(OH)}}_{\text{2}}{}_{\text{(s)}}\\ \text{0}{\text{.125 moles CaC}}_{\text{2}}\text{×}\frac{{\text{1 mole C}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\text{(g)}}}{{\text{1 mole CaC}}_{\text{2}}}\text{ = 0}{\text{.125 moles C}}_{\text{2}}{\text{H}}_{\text{2}}{}_{\text{(g)}}\end{array}$