Question

A method for eliminating oxides of nitrogen (e.g., \(\mathrm{NO}_{2}\) ) from automobile exhaust gases is to pass the exhaust gases over solid cyanuric acid, \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) When the hot exhaust gases come in contact with cyanuric acid, solid \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) decomposes into isocyanic acid vapor, HNCO(g), which then reacts with \(\mathrm{NO}_{2}\) in the exhaust gases to give \(\mathrm{N}_{2}, \mathrm{CO}_{2^{\prime}}\) and \(\mathrm{H}_{2} \mathrm{O}\) How many grams of \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) are needed per gram of \(\mathrm{NO}_{2}\) in this method? [Hint: To balance the equation for reaction between HNCO and \(\mathrm{NO}_{2}\), balance with respect to each kind of atom in this order: \(\mathrm{H}, \mathrm{C}, \mathrm{O}, \text { and } \mathrm{N} .]\)

Short answer

Approximately 0.913 grams of Cyanuric acid (\(C_3N_3(OH)_3\)) are needed to eliminate each gram of Nitric Oxide (\(NO_2\)) in the automobile exhaust.

Step-by-step solution

Write and Balance the Chemical Reaction

Following the hint given in the exercise, start with Hydrogen (H), then Carbon (C), Oxygen (O), and Nitrogen (N) to get the balanced reaction: \[HNCO(g) + NO_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(g).\]

Calculate the Molar Mass of C3N3(OH)3 and NO2

Using the periodic table, calculate the molar mass of Cyanuric acid, \(C_3N_3(OH)_3\), and Nitric oxide, \(NO_2\). The molar mass of \(C_3N_3(OH)_3\) is \(3(12.01g/mol) + 3(14.01g/mol) + 9(1.01g/mol) = 126.07g/mol\). The molar mass of \(NO_2\) is \(14.01g/mol + 2(16.00g/mol) = 46.01g/mol\).

Employing Stoichiometry

The stoichiometric ratio from the balanced equation tells us that one mole of HNCO reacts with one mole of NO2. Because there is only one N atom in NO2 and three N atoms in C3N3(OH)3, we know that three moles of NO2 react with one mole of C3N3(OH)3. Therefore, the mass ratio between Cyanuric acid and Nitric Oxide is \(126.07g/mol \div 3(46.01g/mol) = 0.913g/g\). Hence, approximately 0.913g of Cyanuric acid is required for each gram of Nitric Oxide.

Key concepts

Chemical Reaction Balancing

Understanding how to balance chemical equations is crucial for solving many problems in chemistry. Balancing is done to obey the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction.

Therefore, the same number of each type of atom must appear on both sides of the equation. In the specific exercise given, the task is to balance a reaction involving HNCO and NO₂. Following the hint to balance hydrogen (H) first, then carbon (C), oxygen (O), and nitrogen (N), the correct balanced equation is discovered to be:
\[HNCO(g) + NO_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(g).\]
This step ensures that the atoms involved in the reactants side are equal to those on the products side, allowing for correct stoichiometric calculations later on.

Molar Mass Calculation

When dealing with chemical reactions, molar mass serves as a bridge between the mass of a substance and the amount of substance in moles. Molar mass is defined as the mass of one mole of a substance and is expressed in units of grams per mole (g/mol).

To calculate the molar mass, one must sum the masses of all the atoms in the molecule based on the atomic weights found on the periodic table. In the context of this exercise, the molar masses of cyanuric acid, C₃N₃(OH)₃, and nitric oxide, NO₂, were calculated.

Molar Mass of C₃N₃(OH)₃

\[Molar \text{ Mass } = 3(12.01\text{ g/mol}) + 3(14.01\text{ g/mol}) + 9(1.01\text{ g/mol}) = 126.07\text{ g/mol}.\]

Molar Mass of NO₂

\[Molar \text{ Mass } = 14.01\text{ g/mol} + 2 \times 16.00\text{ g/mol} = 46.01\text{ g/mol}.\]
These calculations are necessary for using stoichiometry to relate the amount of one chemical to another in a reaction.

Stoichiometric Ratio

Stoichiometry refers to the quantitative relationship between the amounts of reactants and products in a chemical reaction. The stoichiometric ratio is derived from a balanced chemical equation and indicates the proportions of reactants and products involved.

In the given exercise, the stoichiometry is straightforward because the balanced equation suggests a 1:1 molar ratio between HNCO and NO₂. However, because cyanuric acid decomposes to form HNCO, the actual stoichiometry involves the C₃N₃(OH)₃:NO₂ ratio. Since a single cyanuric acid molecule provides enough nitrogen for three molecules of NO₂, we see that one mole of C₃N₃(OH)₃ reacts with three moles of NO₂. The mass ratio we need to consider is therefore calculated from the molar masses:
\[Mass \text{ ratio } = 126.07\text{ g/mol } \text{ of } C_3N_3(OH)_3 \text{ to } 3(46.01)\text{ g/mol of } NO_2 = 0.913\text{ g/g}.\]
This ratio tells us how many grams of cyanuric acid are needed to react with each gram of nitric oxide, which is essential for practical applications, such as treating automobile exhaust gases.