Chapter F, Problem L.32E

(a)

Interpretation Introduction

Interpretation:

The number of carbon atoms that is required to react with $\text{600}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ formula units has to be given.

Answer to Problem L.32E

The number of carbon atoms required is $7.2\times {10}^{26}\text{atoms}$.

Explanation of Solution

Reduction reaction given in problem statement is shown below;

    $\begin{array}{c}\text{2C(s)}+{\text{O}}_{\text{2}}\text{(g)}\to 2\text{CO(g)}\\ {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}+3\text{CO(g)}\to 2\text{Fe(l)}\text{+}{\text{3CO}}_{\text{2}}\text{(g)}\end{array}$

The overall reaction can be found by adding the two equations.  The chemical equation obtained is shown below;

    $\text{2C(s)}+{\text{O}}_{\text{2}}{\text{(g) + Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}+\text{CO(g)}\to 2\text{Fe(l)}\text{+}{\text{3CO}}_{\text{2}}\text{(g)}$

The stoichiometric relation between ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ and carbon is that two mol of carbon is required for one mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$.  One formula unit of any substance contains $\text{6}{\text{.023×10}}^{\text{23}}\text{atoms}$.  Therefore, the number of carbon atoms that is required for $\text{600}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ formula units is calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{C}\text{atoms}=\text{600}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{Formula}\text{units}\times \frac{\text{2}\text{mol}\text{C}}{\text{1}\text{mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{formula}\text{units}}\\ \times \frac{\text{6}{\text{.023×10}}^{\text{23}}\text{atoms}}{\text{1}\text{mol}}\\ =7227.6\times {10}^{23}\text{C}\text{atoms}\\ =7.2\times {10}^{26}\text{C}\text{atoms}\end{array}$

Thus the number of carbon atoms that is required by $\text{600}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ formula units is $7.2\times {10}^{26}\text{C}\text{atoms}$.

(b)

Interpretation Introduction

Interpretation:

Maximum amount of carbon dioxide that is generated when $\text{1}\text{.0}\text{t}$ of iron is produced has to be calculated.

Answer to Problem L.32E

Amount of carbon dioxide generated is $9.5\times {10}^5\text{L}$.

Explanation of Solution

The mass of iron is given as $\text{1}\text{.0}\text{t}$.  Therefore, the number of moles in $\text{1}\text{.0}\text{t}$ of iron is calculated as shown below;

    $\begin{array}{c}{n}_{\text{Fe}}={10}^3\text{kg}\text{Fe}\times \frac{\text{1}\text{mol}\text{Fe}}{\text{55}{\text{.845×10}}^{\text{-3}}\text{kg}}\\ =0.017906\times {10}^6\text{mol}\end{array}$

The chemical reaction is given as follows;

    ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}+3\text{CO(g)}\to 2\text{Fe(l)}\text{+}{\text{3CO}}_{\text{2}}\text{(g)}$

The stoichiometric relation between $\text{Fe}$ and carbon dioxide is that two mol of iron is equal to three moles of ${\text{CO}}_{\text{2}}$.  Therefore, the mass of carbon dioxide is calculated as shown below;

    $\begin{array}{c}{m}_{{\text{CO}}_{\text{2}}}=0.017906\times {10}^6\text{mol}\times \frac{\text{3}\text{mol}{\text{CO}}_{\text{2}}}{\text{2}\text{mol}\text{Fe}}\times \frac{\text{44}\text{.01}\text{g}{\text{CO}}_{\text{2}}}{\text{1}\text{mol}{\text{CO}}_{\text{2}}}\\ =1.18\times {10}^6\text{g}\end{array}$

The volume of carbon dioxide is calculated using the density-volume relationship as follows;

    $\begin{array}{c}\text{Volume}=\frac{\text{Mass}}{\text{Density}}\\ =\frac{1.18\times {10}^6\text{g}}{1.25\text{g}\cdot {\text{L}}^{-1}}\\ =0.944\times {10}^6\text{L}\\ =9.5\times {10}^5\text{L}\end{array}$

Thus the volume of carbon dioxide produced is $9.5\times {10}^5\text{L}$.

(c)

Interpretation Introduction

Interpretation:

Maximum amount of carbon dioxide that is released into atmosphere when $\text{1}\text{.0}\text{t}$ of iron is produced with $67.9\%$ yield has to be calculated.

Answer to Problem L.32E

Amount of carbon dioxide released into atmosphere is $6.4\times {10}^5\text{L}$.

Explanation of Solution

The mass of iron is given as $\text{1}\text{.0}\text{t}$.  Yield is given as $67.9\%$.  Therefore, the mass of iron is calculated as follows;

    $\begin{array}{c}{m}_{\text{Fe}}=1\times {10}^3\text{kg}\times \frac{\text{67}\text{.9}}{\text{100}}\\ =0.68\times {10}^3\text{kg}\end{array}$

The chemical reaction is given as follows;

    ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}+3\text{CO(g)}\to 2\text{Fe(l)}\text{+}{\text{3CO}}_{\text{2}}\text{(g)}$

The stoichiometric relation between $\text{Fe}$ and carbon dioxide is that two mol of iron is equal to three moles of ${\text{CO}}_{\text{2}}$.  Therefore, the mass of carbon dioxide is calculated as shown below;

    $\begin{array}{c}{m}_{{\text{CO}}_{\text{2}}}=0.68\times {10}^6\text{g}\text{Fe}\times \frac{\text{1}\text{mol}\text{Fe}}{\text{55}\text{.845}\text{g}\text{Fe}}\times \frac{\text{3}\text{mol}{\text{CO}}_{\text{2}}}{\text{2}\text{mol}\text{Fe}}\times \frac{\text{44}\text{.01}\text{g}{\text{CO}}_{\text{2}}}{\text{1}\text{mol}{\text{CO}}_{\text{2}}}\\ =0.80\times {10}^6\text{g}\end{array}$

The volume of carbon dioxide is calculated using the density-volume relationship as follows;

    $\begin{array}{c}\text{Volume}=\frac{\text{Mass}}{\text{Density}}\\ =\frac{0.80\times {10}^6\text{g}}{1.25\text{g}\cdot {\text{L}}^{-1}}\\ =0.64\times {10}^6\text{L}\\ =6.4\times {10}^5\text{L}\end{array}$

Thus the volume of carbon dioxide released is $6.4\times {10}^5\text{L}$.

(d)

Interpretation Introduction

Interpretation:

Mass of oxygen that is required to produce $\text{5}\text{.00}\text{kg}$ of $\text{Fe}$ has to be calculated.

Answer to Problem L.32E

Mass of oxygen required is $\text{1}\text{.43}\text{kg}$.

Explanation of Solution

Reduction reaction given in problem statement is shown below;

    $\begin{array}{c}\text{2C(s)}+{\text{O}}_{\text{2}}\text{(g)}\to 2\text{CO(g)}\\ {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}+3\text{CO(g)}\to 2\text{Fe(l)}\text{+}{\text{3CO}}_{\text{2}}\text{(g)}\end{array}$

The overall reaction can be found by adding the two equations.  The chemical equation obtained is shown below;

    $\text{2C(s)}+{\text{O}}_{\text{2}}{\text{(g) + Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}+\text{CO(g)}\to 2\text{Fe(l)}\text{+}{\text{3CO}}_{\text{2}}\text{(g)}$

The stoichiometric relation between $\text{Fe}$ and oxygen is that one mole of oxygen molecule is required for two moles of $\text{Fe}$.  Therefore, the moles of Iron in $\text{5}\text{.00}\text{kg}$ is calculated as follows;

    $\begin{array}{c}{n}_{\text{Fe}}=5.00\text{kg}\text{Fe}\times \frac{\text{1}\text{mol}\text{Fe}}{\text{55}{\text{.845×10}}^{\text{-3}}\text{kg}\text{Fe}}\\ =0.08953\times {10}^3\text{mol}\\ =89.5\text{mol}\end{array}$

The mass of oxygen required is calculated as shown below;

    $\begin{array}{c}{m}_{{\text{O}}_{\text{2}}}=89.5\text{mol}{\text{O}}_{\text{2}}\times \frac{\text{1}\text{mol}{\text{O}}_{\text{2}}}{\text{2}\text{mol}\text{Fe}}\times \frac{\text{32}\text{.00}\text{g}{\text{O}}_{\text{2}}}{\text{1}\text{mol}{\text{O}}_{\text{2}}}\\ =1432\text{g}\\ \text{=1}\text{.43}\text{kg}\end{array}$

Thus the mass of oxygen required is $\text{1}\text{.43}\text{kg}$.