91影视

The reaction for complete combustion of gasoline is,

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

The reaction for incomplete combustion of gasoline is,

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

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,

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

Molecules of CO are,

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

So, the ratio of CO₂ to CO is,

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

Thus, the ratio is 19.

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

Mass of CO₂ is,

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

Molecules of CO are,

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

So, the mass ratio of CO₂ to CO is,

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

Thus, the mass ratio is 30.

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

The ratio is,

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

Moles of the compounds are,

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

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,

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

Thus, the percentage of gasoline is 61.1%.