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 g of octane.

We need to write balanced chemical equations for two reactions: 1. Nitrogen reacting with oxygen to produce nitric oxide gas: \[N_2 + O_2 \rightarrow 2NO\] 2. Nitric oxide gas reacting with oxygen to produce nitrogen dioxide gas: \[2NO + O_2 \rightarrow 2NO_2\]

To convert 19 million tons of nitrogen dioxide into grams, we need to use the following conversion factors: 1 ton = 1,000,000 grams So, 19 million tons converted to grams is: \(19 \times 10^6 \, tons \times \frac{1 \times 10^6 \, g}{1 \, ton} = 1.9 \times 10^{13} g\) Thus, 19 million tons of nitrogen dioxide is equal to \(1.9 \times 10^{13}\) grams.

First, we need to determine how much oxygen is required to combust 500 g of octane. The balanced equation for the combustion of octane is: \[C_8H_{18} + \dfrac{25}{2} O_2 \rightarrow 8CO_2 + 9H_2O \] From this equation, we can determine how much oxygen is required for 500 g of octane: 500 g of \(C_8H_{18}\) = \(\dfrac{500 \times \dfrac{25}{2} \, moles \, O_2}{1 \, mole \, C_8H_{18}}\), considering the molar mass \(M(C_8H_{18}) = 114.23 \, g/mol\), and \(M(O_2) = 32 \, g/mol\). which gives: \[\dfrac{500 \times \dfrac{25}{2} \times 32}{114.23} \approx 1093 \, g\, of \, O_2\] Now, considering that 15% of the oxygen produces nitrogen dioxide: \(0.15 \times 1093 = 163.95 \, g\, of \, O_2\) involved in the production of nitrogen dioxide. Using the balanced equation of nitric oxide gas reacting with oxygen: \(2NO + O_2 \rightarrow 2NO_2\) we can calculate the grams of nitrogen dioxide produced: \(163.95 \, g \, O_2 = \dfrac{163.95 \times 2 \, moles \, NO_2}{1 \, mole \, O_2}\), considering the molar mass of \(NO_2 = 46.01 \, g/mol\). which gives: \[\dfrac{163.95 \times 2 \times 46.01}{32} \approx 471.61 \, g\, of \, NO_2\] So, during the combustion of 500 g of octane, 471.61 grams of nitrogen dioxide would be produced.