91Ó°ÊÓ

The reaction for complete combustion of gasoline is,

$2C_8{H}_{18}\left(l\right)+25O_2\left(g\right)→16C{O}_2\left(g\right)+18H_{2}O\left(l\right) ...............\left(1\right)$

The reaction for incomplete combustion of gasoline is,

$2C_8{H}_{18}\left(l\right)+17O_2\left(g\right)→16CO\left(g\right)+18H_{2}O\left(l\right) ...............\left(2\right)$

Let us take 100.0 g solution in which 5.0 g of gasoline is present.

From equation (1) and (2) 2 moles of C₈H₁₈ produces 16 moles of CO₂.

Molecules of CO₂ are,

$$\begin{gathered} =95.0g×\sqrt{b^2-4ac}\frac{1mol C_8{H}_{18}}{114.22g}×\frac{16mol C{O}_2}{2mol C_8{H}_{18}}×\frac{6.022×{10}^{23}molecules}{1mol C{O}_2}\backslash \mathrm{hfill} \\ =4.01×{10}^{24}molecules C{O}_2 \end{gathered}$$

Molecules of CO are,

$$\begin{gathered} =5.0g×\frac{1mol C_8{H}_{18}}{114.22g}×\frac{16mol CO}{2mol C_8{H}_{18}}×\frac{6.022×{10}^{23}molecules}{1mol CO} \\ =2.11×{10}^{23}molecules CO \end{gathered}$$

So, the ratio of CO₂ to CO is,

$$\begin{gathered} =\frac{4.01×{10}^{24}molecules C{O}_2}{2.11×{10}^{23}molecules CO} \\ =19 \end{gathered}$$

Thus, the ratio is 19.

From equation (1) and (2) 2 moles of C₈H₁₈ produces 16 moles of CO₂.

Mass of CO₂ is,

$$\begin{gathered} =95.0g×\frac{1mol C_8{H}_{18}}{114.22g}×\frac{16mol C{O}_2}{2mol C_8{H}_{18}}×\frac{44.01g C{O}_2}{1mol C{O}_2} \\ =293g C{O}_2 \end{gathered}$$

Molecules of CO are,

$$\begin{gathered} =5.0g×\frac{1mol C_8{H}_{18}}{114.22g}×\frac{16mol CO}{2mol C_8{H}_{18}}×\frac{44.01g C{O}_2}{1mol C{O}_2} \\ ⟨\underset{i=1}{\overset{n}{∑}}{X}_i{Y}_i=9.81g CO \end{gathered}$$

So, the mass ratio of CO₂ to CO is,

$$\begin{gathered} =\frac{293g C{O}_2}{9.81g CO} \\ =30 \end{gathered}$$

Thus, the mass ratio is 30.

Let the mass percent of gasoline convert into CO be x.

The ratio is,

$$\begin{gathered} Mass of C{O}_2:Mass of CO  \\ 1.0g:1.0g \end{gathered}$$

Moles of the compounds are,

$$\begin{gathered} \frac{1.0g C{O}_2}{44.01g/mol}:\frac{1.0g CO}{28.01g/mol} \\ 0.02272mol:0.03571mol \end{gathered}$$

Conversion of moles of CO₂ and CO to the number of moles of gasoline

$$\begin{gathered} \frac{0.02272mol}{0.03571mol}:\frac{0.03571mol}{0.03571mol} \\ 0.636:1.0 \end{gathered}$$

Now, the overall moles are,

$$\begin{gathered} =1.0mol+0.636mol \\ =1.636mol \end{gathered}$$

The mass percentage of gasoline is,

$$\begin{gathered} =\frac{1mol C_8{H}_{18}×114.22g/mol}{1.636mol C_8{H}_{18}×114.22g/mol}×100 \\ =61.1\% \end{gathered}$$

Thus, the percentage of gasoline is 61.1%.