The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, principally $\mathrm{N}_{2}(\sim 79 \%)\( and \)\mathrm{O}_{2}(\sim 20 \%) .$ In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As NO is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reactions. (b) Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called "NO \(_{x}\) " gases. In 2009 , the United States emitted an estimated 19 million tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of \(\mathrm{NO}_{x}\) gases is an unwanted side reaction of the main engine combustion process that turns octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) into \(\mathrm{CO}_{2}\) and water. If \(85 \%\) of the oxygen in an engine is used to combust octane and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of \(500 \mathrm{~g}\) of octane.

To write the balanced chemical equations, we need to ensure that the number of atoms of each element on both sides of the equation is equal. For the formation of nitric oxide (NO): \[ N_2 + O_2 \rightarrow 2NO \] For the formation of nitrogen dioxide (NO₂): \[ 2NO + O_2 \rightarrow 2NO_2 \]

We are given that the United States emitted 19 million tons of nitrogen dioxide (NO₂) in 2009. To convert this amount to grams, first, we need to know the number of grams in a ton: \[ 1 \, \text{ton} = 907,184.74 \, \text{grams} \] Then, multiplying the number of tons by the number of grams per ton: \[ 19 \times 10^6 \, \text{tons} \times 907,184.74 \, \frac{\text{grams}}{\text{ton}} \approx 1.72 \times 10^{13} \, \text{grams} \] Hence, 19 million tons of nitrogen dioxide (NO₂) is approximately equal to \(1.72 \times 10^{13}\) grams.

We are given that 85% of the oxygen in an engine is used to combust octane and the remaining 15% is used to produce nitrogen dioxide. We are also given that 500 grams of octane are combusted. To calculate the amount of nitrogen dioxide produced, we will follow these steps: 1. Write the balanced chemical equation for the combustion of octane: \[ 2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O \] 2. Determine the number of moles of octane combusted using its molar mass (114.23 g/mol): \[ \frac{500 \, \text{g}}{114.23 \, \frac{\text{g}}{\text{mol}}} \approx 4.374 \, \text{moles} \] 3. Calculate the number of moles of oxygen used for combustion from the balanced chemical equation: \[ \frac{4.374 \, \text{moles} \times 25}{2} \approx 54.68 \, \text{moles} \] 4. Calculate the total moles of oxygen before the combustion, considering that 85% of it was used for combustion: \[ \frac{54.68 \, \text{moles}}{0.85} \approx 64.33 \, \text{moles} \] 5. Find the moles of oxygen used for producing nitrogen dioxide (the remaining 15%): \[ 0.15 \times 64.33 \, \text{moles} \approx 9.65 \, \text{moles} \] 6. Calculate the moles of nitrogen dioxide produced using the balanced chemical equation: \[ \frac{9.65 \, \text{moles} \times 2}{1} \approx 19.3 \, \text{moles} \] 7. Convert the moles of nitrogen dioxide to grams using its molar mass (46.01 g/mol): \[ 19.3 \, \text{moles} \times 46.01 \, \frac{\text{g}}{\text{mol}} \approx 887.99 \, \text{grams} \] Hence, approximately 887.99 grams of nitrogen dioxide will be produced during the combustion of 500 grams of octane.