Question

Question: In a car engine, gasoline (represented by C8H18) does not burn completely, and some CO, a toxic pollutant, forms along with CO2 and H2O. If 5.0% of the gasoline forms CO:

(a) What is the ratio of CO₂ to CO molecules in the exhaust?

(b) What is the mass ratio of CO₂ to CO?

(c) What percentage of the gasoline must form CO for the mass ratio of CO₂ to CO to be exactly 1/1?

Short answer

Answer:

1. The ratio is 19.
2. The mass ratio is 30.
3. The percentage of gasoline 61.1%.

Step-by-step solution

(a) Determination of ratio of CO2 to CO

The reaction for complete combustion of gasoline is,

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

The reaction for incomplete combustion of gasoline is,

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

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,

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

Molecules of CO are,

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

So, the ratio of CO₂ to CO is,

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

Thus, the ratio is 19.

 Step 2: (b) Determination of mass ratio of CO2 to CO 

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

Mass of CO₂ is,

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

Molecules of CO are,

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

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.

(c) Determination of percentage of gasoline 

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,

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

Thus, the percentage of gasoline is 61.1%.