Chapter 9, Problem 52A

(a)

Interpretation Introduction

Interpretation:

The number of moles of each product e produced by complete conversion of 0.50 moles of the reactant indicated bold in given reaction needs to be determined.

  $2 H_2{O}_2{}_{ \left(l\right)} \to 2H_2{O}_{ \left(l\right)} +O_{2\left(g\right)}$

Concept Introduction:

If complete conversion of any reactant takes place in the chemical reaction, it is considered as a limiting reactant.

Answer to Problem 52A

Moles of H₂O = 0.5 moles H₂O

Moles of O₂ = 0.25 moles O₂

Explanation of Solution

The given reaction is as follows:

  $2 H_2{O}_2{}_{ \left(l\right)} \to 2H_2{O}_{ \left(l\right)} +O_{2\left(g\right)}$

Given: 0.5 moles H₂O₂

Mole ratio H₂O₂ : H₂O = 2:2 = 1:1

Thus,

  $molesof{H}_{2}O=\frac{1}{1}\times 0.5moles=0.5moles{H}_{2}O$

Mole Ratio H₂O₂ ; O₂ = 2:1

Thus,

  $moles{O}_2 =\frac{1}{2}\times 0.5moles=0.25moles{O}_2$

(b)

Interpretation Introduction

Interpretation:

The number of moles of each product e produced by complete conversion of 0.50 moles of the reactant indicated bold in given reaction needs to be determined.

  $2 KCl{O}_{3 }{}_{\left(s\right)} \to 2KCl{ }_{\left(s\right)} +3O_{2\left(g\right)}$

Concept Introduction:

If complete conversion of any reactant takes place in the chemical reaction, it is considered as a limiting reactant.

Answer to Problem 52A

Moles of KCl = 0.5 moles KCl

Moles of O₂ = 0.75 moles O₂

Explanation of Solution

The given reaction is as follows:

  $2 KCl{O}_{3 }{}_{\left(s\right)} \to 2KCl{ }_{\left(s\right)} +3O_{2\left(g\right)}$

Moles of KClO₃ = 0.5 moles

Mole Ratio KClO₃ : KCl = 2:2 = 1:1

Thus,

  $\text{moles KCl=}\frac{1}{1}\times 0.5\text{ mol=0}\text{.5 mol KCl}$

Mole Ratio KClO₃ : O₂ = 2:3

Thus,

  ${\text{moles O}}_{\text{2}}\text{=}\frac{3}{2}\times 0.5\text{ mol=0}{\text{.75 mol O}}_{\text{2}}$

(c)

Interpretation Introduction

Interpretation:

The number of moles of each product produced by complete conversion of 0.50 moles of the reactant indicated in boldface needs to be determined.

  $2 A{l}_{ (s)} +6HC{l}_{ (aq)} \to 2AlC{l}_{3(aq)} +3H_{2(g)}$

Concept Introduction:

If complete conversion of any reactant takes place in the chemical reaction, it is considered as a limiting reactant.

Answer to Problem 52A

Moles of AlCl₃ = 0.5 moles AlCl₃

Moles of H₂ = 0.75 moles H₂

Explanation of Solution

The given reaction is as follows:

  $2 A{l}_{ (s)} +6HC{l}_{ (aq)} \to 2AlC{l}_{3(aq)} +3H_{2(g)}$

Moles of Al = 0.5 moles

Mole Ratio Al : AlCl₃ = 2:2 = 1:1

Thus,

  $MolesofAlC{l}_3 =\frac{1}{1}\times 0.5moles=0.5molesAlC{l}_3$

Mole Ratio Al : H₂ = 2:3

Thus,

  $Molesof H_2 =\frac{3}{2}\times 0.5moles=0.75moles{H}_2$

(d)

Interpretation Introduction

Interpretation:

The number of moles of each product produced by complete conversion of 0.50 moles of the reactant indicated in boldface needs to be determined.

  $C_3{H}_8{}_{ \left(g\right)} +5O_{2\left(g\right)} \to 3C{O}_{2\left(g\right)} +4H_2{O}_{ \left(g\right)}$

Concept Introduction:

If complete conversion of any reactant takes place in the chemical reaction, it is considered as a limiting reactant.

Answer to Problem 52A

moles CO₂ = 1.5 moles CO₂

moles H₂O = 2.0 moles H₂O

Explanation of Solution

  $C_3{H}_8{}_{ \left(g\right)} +5O_{2\left(g\right)} \to 3C{O}_{2\left(g\right)} +4H_2{O}_{ \left(g\right)}$

Moles of C₃H₈ = 0.5 moles

Mole Ratio C₃H₈ : CO₂ = 1:3

Thus,

  ${\text{moles CO}}_{\text{2}}=\frac{3}{1}\times 0.5\text{ mol=1}{\text{.5 mol CO}}_{\text{2}}$

Mole Ratio C₃H₈ : H₂O = 1:4

Thus,

  ${\text{moles H}}_{\text{2}}\text{O}=\frac{4}{1}\times 0.5\text{ mol=2}{\text{.0 mol H}}_{\text{2}}\text{O}$