Chapter 10, Problem 61A

(a)

Interpretation Introduction

Interpretation: The heat released when 4.00 moles of iron is reacted with excess of ${\text{O}}_{\text{2}}$ for the given reaction needs to be determined.

  ${\text{4Fe}}_{\text{(s)}}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}\text{ ΔH = -1652kJ}$

Concept Introduction: On the basis of energy change, a chemical reaction can be classified as exothermic and endothermic reaction.

The change in enthalpy of reaction is always calculated for the balanced chemical equation and can be used to calculate the energy change for the given moles or masses of reactant or product molecules.

Answer to Problem 61A

Heat released with 4.00 moles of Fe = 1652 kJ

Explanation of Solution

Given:

  ${\text{4Fe}}_{\text{(s)}}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}\text{ ΔH = -1652kJ}$

Moles of Fe = 4.00 moles

According to balance chemical equation:

4 moles of Fe released 1652 kJ of energy.

Thus heat released with 4.00 moles of Fe = 1652 kJ

(b)

Interpretation Introduction

Interpretation: The heat released when 1.00 g of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}$ is produced needs to be determined.

  ${\text{4Fe}}_{\text{(s)}}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}\text{ ΔH = -1652kJ}$

Concept Introduction: On the basis of energy change, a chemical reaction can be classified as exothermic and endothermic reaction.

The change in enthalpy of reaction is always calculated for the balanced chemical equation and can be used to calculate the energy change for the given moles or masses of reactant or product molecules.

Answer to Problem 61A

  $\text{ 5}\text{.172 kJ}$

Explanation of Solution

Given:

  ${\text{4Fe}}_{\text{(s)}}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}\text{ ΔH = -1652kJ}$

Mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}$ = 1.00 g

Molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}$ = 159.7 g/mol

Calculate moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}$ :

  ${\text{moles of Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}\text{=}\frac{\text{mass}}{\text{molar mass}}\text{=}\frac{\text{1}\text{.00 g}}{\text{159}\text{.7 g/mol }}\text{= 0}\text{.00626 moles}$

According to balance chemical equation:

2 moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}$ released 1652 kJ of energy.

  $\text{ 0}{\text{.00626 moles Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}\times \frac{\text{1625 kJ}}{{\text{2 moles of Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}}\text{= 5}\text{.172 kJ}$

(c)

Interpretation Introduction

Interpretation: The heat released when 1.00 g of $\text{Fe}$ is produced needs to be determined.

  ${\text{4Fe}}_{\text{(s)}}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}\text{ ΔH = -1652kJ}$

Concept Introduction: On the basis of energy change, a chemical reaction can be classified as exothermic and endothermic reaction.

The change in enthalpy of reaction is always calculated for the balanced chemical equation and can be used to calculate the energy change for the given moles or masses of reactant or product molecules.

Answer to Problem 61A

  $\text{7}\text{.39 kJ}$

Explanation of Solution

Given:

  ${\text{4Fe}}_{\text{(s)}}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}\text{ ΔH = -1652kJ}$

Mass of $\text{Fe}$ = 1.00 g

Molar mass of $\text{Fe}$ = 55.8 g/mol

Calculate moles of $\text{Fe}$ :

  $\text{moles of Fe =}\frac{\text{mass}}{\text{molar mass}}\text{=}\frac{\text{1}\text{.00 g}}{\text{55}\text{.8 g/mol }}\text{= 0}\text{.0179 moles}$

According to balance chemical equation:

4 moles of $\text{Fe}$ released 1652 kJ of energy.

  $\text{ 0}\text{.0179 moles Fe}\times \frac{\text{1652 kJ}}{\text{4 moles of Fe}}\text{= 7}\text{.39 kJ}$

(d)

Interpretation Introduction

Interpretation: The heat released when 10.0 g of $\text{Fe}$ and 2.00 g of ${\text{ O}}_{\text{2}}$ are reacted needs to be determined.

  ${\text{4Fe}}_{\text{(s)}}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}\text{ ΔH = -1652kJ}$

Concept Introduction: On the basis of energy change, a chemical reaction can be classified as exothermic and endothermic reaction.

The change in enthalpy of reaction is always calculated for the balanced chemical equation and can be used to calculate the energy change for the given moles or masses of reactant or product molecules.

Answer to Problem 61A

34.4 kJ energy will release.

Explanation of Solution

Given:

  ${\text{4Fe}}_{\text{(s)}}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}^{}\text{ ΔH = -1652kJ}$

Mass of $\text{Fe}$ = 10.0 g

Mass of ${\text{ O}}_{\text{2}}$ = 2.00 g

Molar mass of $\text{Fe}$ = 55.8 g/mol

Molar mass of ${\text{ O}}_{\text{2}}$ = 32.0 g

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

  $\begin{array}{l}\text{moles of Fe =}\frac{\text{mass}}{\text{molar mass}}\text{=}\frac{\text{10}\text{.0 g}}{\text{55}\text{.8 g/mol }}\text{= 0}\text{.179 moles}\\ {\text{moles of O}}_{\text{2}}\text{ =}\frac{\text{mass}}{\text{molar mass}}\text{=}\frac{\text{2}\text{.00 g}}{\text{32}\text{.0 g/mol }}\text{= 0}\text{.0625 moles}\end{array}$

According to balance chemical equation:

4 moles of $\text{Fe}$ = 3 moles of ${\text{ O}}_{\text{2}}$ = released 1652 kJ of energy.

  $\text{ 0}\text{.179 moles Fe}\times \frac{{\text{3 moles of O}}_{\text{2}}}{\text{4 moles of Fe}}\text{= 0}{\text{.134 moles O}}_{\text{2}}$

Since the calculated moles of ${\text{ O}}_{\text{2}}$ are less therefore it should be limiting reagent and will determine the heat released.

  $\text{ 0}{\text{.0625 moles O}}_{\text{2}}\times \frac{\text{1652 kJ}}{{\text{3 moles of O}}_{\text{2}}}\text{= 34}\text{.4 kJ}$