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 $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathit{g}\mathit{a}\mathit{l}\mathbf{/}\mathit{d}\mathit{a}\mathit{y}$.)

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.

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 –

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

(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 $\text{9.0044}\times {\text{10}}^{\text{9}}\text{g}$.

(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 $\text{4.316}\times {\text{10}}^{\text{10}}\text{g}$.