Chapter 3: Problem 112

\(\mathrm{NO}_{x}\) is a generic term for the nitrogen oxides, \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) \(\mathrm{NO}_{x}\) gases are air pollutants that react to form smog and acid rain. In order to reduce \(\mathrm{NO}_{x}\) emission from vehicle, catalytic converters are installed in car exhausts to decompose NO and \(\mathrm{NO}_{2}\) respectively into \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}(\mathbf{a})\) Write the balanced chemical equations for the decomposition of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) respectively. (b) If the car produces \(100 \mathrm{~g} \mathrm{NO}_{x}\) a day, with equal mole ratio of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\), how many grams of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) are produced respectively?

Short Answer

Expert verified

The balanced chemical equations for the decomposition of NO and NO₂ are: \[2NO \rightarrow N_{2} + O_{2}\] and \[2NO_{2} \rightarrow N_{2} + 2O_{2}\] The car produces \(39.5\mathrm{~g}\) of NO and \(60.5\mathrm{~g}\) of NO₂ daily.

Step by step solution

The balanced chemical equation for the decomposition of NO is: \[2NO \rightarrow N_{2} + O_{2}\] #a2. Write the decomposition equation for NO₂# To write the balanced chemical equation for the decomposition of NO₂, we need to make sure the number of atoms for each element is equal on both sides of the equation. NO₂ decomposes into N₂ and O₂.

The balanced chemical equation for the decomposition of NO₂ is: \[2NO_{2} \rightarrow N_{2} + 2O_{2}\] #b1. Calculate moles of NOₓ produced daily# To calculate the moles of NOₓ produced daily, we first need to find the molar mass of NOₓ. Since we know that equal mole ratio of NO and NO₂ is produced daily, we can calculate the average molar mass of NOₓ.

The molar mass of NO is \(30.01\mathrm{g/mol}\) (14.01 for N and 16.00 for O) and the molar mass of NO₂ is \(46.01\mathrm{g/mol}\) (14.01 for N and 32.00 for O₂). The average molar mass of NOₓ is: \[\frac{30.01 + 46.01 }{2} = 38.01\mathrm{g/mol}\] Now, we can find the moles of NOₓ produced daily: \[\frac{100\mathrm{~g}}{38.01\mathrm{g/mol}} = 2.63\mathrm{~mol}\] #b2. Calculate the grams of NO and NO₂ produced daily# Since we know the equal mole ratio of NO and NO₂, we can now find the grams of each produced daily.

Divide the moles of NOₓ by 2 to find the moles of NO and NO₂ produced daily: \[2.63\mathrm{~mol}/2 = 1.315\mathrm{~mol}\] Now, multiply the moles of NO and NO₂ by their respective molar masses to find the grams of each produced daily: \[1.315\mathrm{~mol} \times 30.01\mathrm{g/mol} = 39.5\mathrm{~g}\,NO\] \[1.315\mathrm{~mol} \times 46.01\mathrm{g/mol} = 60.5\mathrm{~g}\,NO_{2}\] Hence, the car produces \(39.5\mathrm{~g}\) of NO and \(60.5\mathrm{~g}\) of NO₂ daily.