Chapter 9, Problem 55A

(a)

Interpretation Introduction

Interpretation: The number of moles of product formed with 0.50 moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ in the given chemical reaction needs to be determined.

  ${\text{2 H}}_{\text{2}}{\text{O}}_{\text{2(l)}}\stackrel{}{\to }{\text{2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + O}}_{\text{2(g)}}$

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. It is used to determine the amount of product formed during the chemical reaction.

Answer to Problem 55A

  $\text{ 0}{\text{.50 moles H}}_{\text{2}}\text{O and 0}{\text{.25 moles O}}_{\text{2}}$

Explanation of Solution

The balance chemical reaction:

  ${\text{2 H}}_{\text{2}}{\text{O}}_{\text{2(l)}}\stackrel{}{\to }{\text{2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + O}}_{\text{2(g)}}$

Thus 2 mole of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ reacts = 2 moles of ${\text{H}}_{\text{2}}\text{O}$ = 1 mole of ${\text{O}}_{\text{2}}$

Mole of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ = 0.50 moles

Calculate moles of each product:

  $\begin{array}{l}\text{0}{\text{.50 moles H}}_{\text{2}}{\text{O}}_{\text{2}}\text{ × }\frac{{\text{2 moles H}}_{\text{2}}\text{O}}{{\text{2 moles H}}_{\text{2}}{\text{O}}_{\text{2}}}\text{= 0}{\text{.50 moles H}}_{\text{2}}\text{O}\\ \text{0}{\text{.50 moles H}}_{\text{2}}{\text{O}}_{\text{2}}\text{ × }\frac{{\text{1 moles O}}_{\text{2}}}{{\text{2 moles H}}_{\text{2}}{\text{O}}_{\text{2}}}\text{= 0}{\text{.25 moles O}}_{\text{2}}\end{array}$

(b)

Interpretation Introduction

Interpretation: The number of moles of product formed with 0.50 moles of ${\text{KClO}}_{\text{3}}$ in the given chemical reaction needs to be determined.

  ${\text{2 KClO}}_{\text{3}}{}_{\text{(s)}}\stackrel{}{\to }{\text{2 KCl}}_{\text{(s)}}{\text{ + 3 O}}_{\text{2(g)}}$

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. It is used to determine the amount of product formed during the chemical reaction.

Answer to Problem 55A

  $\text{0}\text{.25 moles KCl and 0}{\text{.75 moles O}}_{\text{2}}$

Explanation of Solution

The balance chemical reaction:

  ${\text{2 KClO}}_{\text{3}}{}_{\text{(s)}}\stackrel{}{\to }{\text{2 KCl}}_{\text{(s)}}{\text{ + 3 O}}_{\text{2(g)}}$

Thus 2 mole of ${\text{KClO}}_{\text{3}}$ reacts = 2 moles of $\text{KCl}$ = 3 mole of ${\text{O}}_{\text{2}}$

Mole of ${\text{KClO}}_{\text{3}}$ = 0.50 moles

Calculate moles of each product:

  $\begin{array}{l}\text{0}{\text{.50 moles KClO}}_{\text{3}}\text{× }\frac{\text{1 moles KCl}}{{\text{2 moles KClO}}_{\text{3}}}\text{= 0}\text{.25 moles KCl}\\ \text{0}{\text{.50 moles KClO}}_{\text{3}}\text{ × }\frac{{\text{3 moles O}}_{\text{2}}}{{\text{2 moles KClO}}_{\text{3}}}\text{= 0}{\text{.75 moles O}}_{\text{2}}\end{array}$

(c)

Interpretation Introduction

Interpretation: The number of moles of product formed with 0.50 moles of $\text{Al}$ in the given chemical reaction needs to be determined.

  ${\text{2 Al}}_{\text{(s)}}{\text{ + 6 HCl}}_{\text{(aq)}}\stackrel{}{\to }{\text{2 AlCl}}_{\text{3}}{}_{\text{(aq)}}{\text{ + 3 H}}_{\text{2(g)}}$

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. It is used to determine the amount of product formed during the chemical reaction.

Answer to Problem 55A

  $\text{0}{\text{.25 moles AlCl}}_{\text{3}}\text{ and 0}{\text{.75 moles H}}_{\text{2}}$

Explanation of Solution

The balance chemical reaction:

  ${\text{2 Al}}_{\text{(s)}}{\text{ + 6 HCl}}_{\text{(aq)}}\stackrel{}{\to }{\text{2 AlCl}}_{\text{3}}{}_{\text{(aq)}}{\text{ + 3 H}}_{\text{2(g)}}$

Thus 2 mole of Al reacts = 2 moles of ${\text{AlCl}}_{\text{3}}$ = 3 mole of ${\text{H}}_{\text{2(g)}}$

Mole of Al = 0.50 moles

Calculate moles of each product:

  $\begin{array}{l}\text{0}\text{.50 moles Al × }\frac{{\text{1 moles AlCl}}_{\text{3}}}{\text{2 moles Al}}\text{= 0}{\text{.25 moles AlCl}}_{\text{3}}\\ \text{0}\text{.50 moles Al × }\frac{{\text{3 moles H}}_{\text{2}}}{\text{2 moles Al}}\text{ = 0}{\text{.75 moles H}}_{\text{2}}\end{array}$

(d)

Interpretation Introduction

Interpretation: The number of moles of product formed with 0.50 moles of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ in the given chemical reaction needs to be determined.

  ${\text{2C}}_{\text{3}}{\text{H}}_{\text{8}}{}_{\text{(g)}}{\text{ + 5 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{3 CO}}_{\text{2(g)}}{}_{\text{(aq)}}{\text{ + 4H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

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. It is used to determine the amount of product formed during the chemical reaction.

Answer to Problem 55A

  $\text{0}{\text{.75 moles CO}}_{\text{2}}\text{ and 1}{\text{.00 moles H}}_{\text{2}}\text{O}$

Explanation of Solution

The balance chemical reaction:

  ${\text{2C}}_{\text{3}}{\text{H}}_{\text{8}}{}_{\text{(g)}}{\text{ + 5 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{3 CO}}_{\text{2(g)}}{}_{\text{(aq)}}{\text{ + 4H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Thus 2 mole of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ reacts = 3 moles of ${\text{CO}}_{\text{2}}$ = 4 mole of ${\text{H}}_{\text{2}}\text{O}$

Mole of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ = 0.50 moles

Calculate moles of each product:

  $\begin{array}{l}\text{0}{\text{.50 moles C}}_{\text{3}}{\text{H}}_{\text{8}}\text{ × }\frac{{\text{3 moles CO}}_{\text{2}}}{{\text{2 moles C}}_{\text{3}}{\text{H}}_{\text{8}}}\text{= 0}{\text{.75 moles CO}}_{\text{2}}\\ \text{0}{\text{.50 moles C}}_{\text{3}}{\text{H}}_{\text{8}}\text{ × }\frac{{\text{4 moles H}}_{\text{2}}\text{O}}{{\text{2 moles C}}_{\text{3}}{\text{H}}_{\text{8}}}\text{ = 1}{\text{.00 moles H}}_{\text{2}}\text{O}\end{array}$