Chapter 10, Problem 63A

(a)

Interpretation Introduction

Interpretation: The heat released due to reaction of 4.00 moles iron with excess ${\text{O}}_{\text{2}}$ needs to be calculated.

Concept Introduction: The energy flow between two objects due to difference of temperature is termed as heat. In exothermic process, heat is released from system to surroundings and in endothermic process absorption of heat by the system from surroundings occurs.

Answer to Problem 63A

Reaction of 4.00 moles of iron with excess ${\text{O}}_{\text{2}}$ releases heat of $1652\text{kJ}$ .

Explanation of Solution

The given reaction is as follows:

  $4\text{Fe(}s\text{)}{\text{+3O}}_2(g)\to 2{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}(s)\Delta H=-1652\text{kJ}$

The given reaction indicates that 4 moles of $\text{Fe}$ reacts with 3 moles of ${\text{O}}_{\text{2}}$ to release $1652\text{kJ}$ of heat. Thus, heat produced on reaction of 4 moles of Fe is with excess ${\text{O}}_{\text{2}}$ is $1652\text{kJ}$ .

(b)

Interpretation Introduction

Interpretation: The quantity of heat released on production of 1.00 mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ needs to be calculated.

Concept Introduction: The energy flow between two objects due to difference of temperature is termed as heat. In exothermic process, heat is released from system to surroundings and in endothermic process absorption of heat by the system from surroundings occurs.

Answer to Problem 63A

The production of $\text{1}.00\text{ mole}$ of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ releases heat of $826\text{kJ}$ of.

Explanation of Solution

The given reaction is as follows:

  $4\text{Fe(}s\text{)}{\text{+3O}}_2(g)\to 2{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}(s)\Delta H=-1652\text{kJ}$

When 2 mol of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ produced $1652\text{kJ}$ of energy released. So, energy released in formation of one mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ can be calculated as,

  $\begin{array}{l}=1\text{mol}{\text{Fe}}_2{\text{O}}_{\text{3}}\times \frac{1652\text{kJ}}{2\text{mol}{\text{Fe}}_2{\text{O}}_{\text{3}}}\\ =826\text{kJ}\end{array}$

(c)

Interpretation Introduction

Interpretation: The amount of heat released due to reaction of 1.00 g of iron with excess ${\text{O}}_{\text{2}}$ needs to be calculated.

Concept Introduction: The energy flow between two objects due to difference of temperature is termed as heat. In exothermic process, heat is released from system to surroundings and in endothermic process absorption of heat by the system from surroundings occurs.

Answer to Problem 63A

The reaction of 1.00 g of iron is reacted with excess ${\text{O}}_{\text{2}}$ releases $7.39\text{kJ}$ of heat.

Explanation of Solution

First calculate the moles of iron in 1.00 g .

Molar mass of iron is $55.84\text{g/mol}$ .

  $\begin{array}{c}\text{Moles=}\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{1.00\text{g}}{55.84\text{g/mol}}\\ =0.0179\text{mol}\end{array}$

Now calculate the amount of heat releases from $0.0179\text{mol}$ of $\text{Fe}$ .

  $\begin{array}{c}\text{Heat}\text{releases}=\text{0}\text{.0179}\text{mol}\text{Fe}\times \frac{1652\text{kJ}}{4\text{mol}\text{Fe}}\\ =7.39\text{kJ}\end{array}$

(d)

Interpretation Introduction

Interpretation: The quantity of heat released due to reaction of $\text{1}0.0\text{ g}$ of $\text{Fe}$ and $2.00\text{g}$ of ${\text{O}}_{\text{2}}$ needs to be calculated.

Concept Introduction: The energy flow between two objects due to difference of temperature is termed as heat. In exothermic process, heat is released from system to surroundings and in endothermic process absorption of heat by the system from surroundings occurs.

Answer to Problem 63A

Heat released due to reaction of 10.0 of $\text{Fe}$ and $2.00\text{g}$ of ${\text{O}}_{\text{2}}$ is $34.4\text{kJ}$ .

Explanation of Solution

First calculate the moles of iron in $10\text{g}$ of iron.

Molar mass of iron is $55.84\text{g/mol}$ .

  $\begin{array}{c}\text{Moles=}\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{10\text{g}}{55.84\text{g/mol}}\\ =0.179\text{mol}\end{array}$

Then, calculate the moles of oxygen in $2\text{g}$ of oxygen.

Molar mass of iron is $32.0\text{g/mol}$ .

  $\begin{array}{c}\text{Moles=}\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{2\text{g}}{32.0\text{g/mol}}\\ =0.0625\text{mol}\end{array}$

Now calculate the moles of oxygen needed to react with $0.179\text{mol}$ of Fe.

  $\begin{array}{c}=0.179\text{mol}\text{Fe}\times \frac{3\text{mol}{\text{O}}_2}{4\text{mol}\text{Fe}}\\ =0.134\text{mol}{\text{O}}_2\end{array}$

Amount of oxygen needed is greater than the amount of oxygen present. Thus, the limiting reactant is oxygen.

Now calculate the heat released from 0.0625 mol oxygen.

  $\begin{array}{c}=0.0625\text{mol}{\text{O}}_2\times \frac{1652\text{kJ}}{3\text{mol}{\text{O}}_2}\\ =34.4\text{kJ}\end{array}$