Question

A blast furnace uses ${\mathbf{\text{Fe}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{3}}}$ to produce$\mathbf{\text{8400 t}}$ of $\mathbf{\text{Fe}}$per day.

(a) What mass of ${\mathbf{\text{CO}}}_{\mathbf{\text{2}}}$is produced each day?

(b) Compare this amount of ${\text{CO}}_{\text{2}}$with that produced by1.0million automobiles, each burning role="math" localid="1663324208860" $\mathbf{\text{5.0 gal}}$of gasoline a day. Assume that gasoline has the formula${\mathbf{\text{C}}}_{\mathbf{\text{8}}}{\mathbf{\text{H}}}_{\mathbf{\text{18}}}$ and a density of $\mathbf{\text{0.74 g/mL}}$, and that it burns completely. (Note that U.S. gasoline consumption is over $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathit{g}\mathit{a}\mathit{l}\mathbf{/}\mathit{d}\mathit{a}\mathit{y}$.)

Short answer

(a) The mass of ${\text{CO}}_{\text{2}}$ produced each day is $\text{9.0044}\times {\text{10}}^{\text{9}}\text{g}$.

(b) The mass of ${\text{CO}}_{\text{2}}$ produced by cars is $\text{4.316}\times {\text{10}}^{\text{10}}\text{g}$. Comparing it with the mass of carbon dioxide produced in (a), $\text{1 million}$ cars produce more ${\text{CO}}_{\text{2}}$.

Step-by-step solution

Concept Introduction

In combustion process a substance is burned in the presence of oxygen, and results in the generation of heat and light as a result.

Combustion is demonstrated by the examples like burning of sulphur in the air and the explosion of hydrogen in the air.

Mass of Iron

Writing the reaction for the blast furnace –

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

Given $\text{8400 t}$- American tons (not the metric ton, $\text{1000 kg}$), convert the $t$ to the SI units.

$$\begin{gathered} 1t=1ton=2000lbs \\ 1lbs=\frac{1}{2205}kg \\ 1kg={10}^{3}gm \end{gathered}$$

Thus, it is obtained that –

$$\begin{gathered} \left(Fe\right)=\frac{8400t\times 2000\frac{lbs}{t}}{\frac{1}{2205}\frac{kg}{lbs}}\times {10}^3\frac{g}{kg} \\ m\left(Fe\right)=7619047619g=7.619\times {10}^{9}g \end{gathered}$$

Mass of Carbon Dioxide

(a)

Based on the reaction, $\text{3 mol}$ of ${\text{CO}}_{\text{2}}$ are produced along with $\text{2 mol}$ of $\text{Fe}$.

Given the mass of $\text{Fe}$, convert it to amount of substance –

$$\begin{gathered} n\left(Fe\right)=\frac{m}{M_R} \\ n\left(Fe\right)=\frac{7.619\times {10}^{9}g}{55.845g/mol} \\ n\left(Fe\right)=1.364\times {10}^{8}mol \end{gathered}$$

Thus, the amount of ${\text{CO}}_{\text{2}}$ produced is –

$$\begin{gathered} n\left(C{O}_2\right)=\frac{3}{2}\times n\left(Fe\right) \\ n\left(C{O}_2\right)=\frac{3}{2}\times 1.364\times {10}^{8}mol \\ n\left(C{O}_2\right)=2.046\times {10}^{8}mol \end{gathered}$$

Given the molar mass of ${\text{CO}}_{\text{2}}$, the mass produced can be calculated –

$$\begin{gathered} m\left(C{O}_2\right)=n\times M_R \\ m\left(C{O}_2\right)=2.046\times {10}^{8}mol\times 44.01g/mol \\ m\left(C{O}_2\right)=9.0044\times {10}^{9}g \end{gathered}$$

Therefore, the value for mass is obtained as $\text{9.0044}\times {\text{10}}^{\text{9}}\text{g}$.

Comparing the mass of Carbon Dioxide

(b)

At first, calculate the gasoline burned by the$\text{1 mil}$automobiles –

$$\begin{gathered} V\left(gasoline\right)=1\times {10}^{6}auto\times 5.0\frac{gal}{day·auto} \\ V\left(gasoline\right)=5\times {10}^6\frac{gal}{day} \end{gathered}$$

Converting gal to Sl units, the gasoline burned per day –

$$\begin{gathered} 1gal=3.785L \\ 1L={10}^{3}mL \\ V\left(gasoline\right)=5\times {10}^{6}gal\times 3.785\frac{L}{gas}\times {10}^3\frac{mL}{L} \\ V\left(gasoline\right)=1.8925\times {10}^{10}mL \end{gathered}$$

Knowing the volume and density of gasoline, its mass and mol number can be calculated –

$$\begin{gathered} m\left(gasoline\right)=d\times V=0.74g/mL\times 1.8925·{10}^{10}mL \\ m\left(gasoline\right)=1.40045\times {10}^{10}g \\ n\left(gasoline\right)=\frac{m}{M_R}=\frac{1.40045\times {10}^{10}g}{114.22g/mol} \\ n\left(gasoline\right)=1.2261\times {10}^{8}mol \end{gathered}$$

The combustion reaction of gasoline is –

${\text{2C}}_{\text{8}}{\text{H}}_{\text{18}}{\text{(l) + 25O}}_{\text{2}}\text{(g)}\to {\text{16CO}}_{\text{2}}{\text{(g) + 18H}}_{\text{2}}\text{O(l)}$

Fromof gasoline, 16 mol of${\text{CO}}_{\text{2}}$are produced.

The amount of${\text{CO}}_{\text{2}}$produced then –

$$\begin{gathered} n\left(C{O}_2\right)=\frac{16}{2}\times n\left(gasoline\right) \\ n\left(C{O}_2\right)=8\times 1.2261\times {10}^{8}mol \\ n\left(C{O}_2\right)=9.8088\times {10}^8\left(mol\right) \end{gathered}$$

Finally, the mass of${\text{CO}}_{\text{2}}$produced by$\text{1 million}$cars –

$$\begin{gathered} m\left(C{O}_2\right)=n·M_R \\ m\left(C{O}_2\right)=9.8088\times {10}^8\left(mol\right)\times 44.01g/mol \\ m\left(C{O}_2\right)=4.316\times {10}^{10}g \end{gathered}$$

Therefore, the value for ${\text{CO}}_{\text{2}}$ produced for $\text{1 mil}$ cars is obtained as $\text{4.316}\times {\text{10}}^{\text{10}}\text{g}$.