Question

Nitrogen oxide is a pollutant commonly found in smokestack emissions. One way to remove it is to react it with ammonia. $$ 4 \mathrm{NH}_{3}(g)+6 \mathrm{NO}(g) \longrightarrow 5 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ How many liters of ammonia are required to change \(12.8 \mathrm{~L}\) of nitrogen oxide to nitrogen gas? Assume \(100 \%\) yield and that all gases are measured at the same temperature and pressure.

Short answer

Answer: 8.533 L of Ammonia are required to react completely with 12.8 L of Nitrogen oxide.

Step-by-step solution

Analyze the balanced equation to understand the stoichiometry

In the balanced equation: $$ 4 \mathrm{NH}_{3}(g)+6 \mathrm{NO}(g) \longrightarrow 5 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ For every 6 moles of Nitrogen oxide, 4 moles of Ammonia are required to react completely.

Use the relationship between volumes and moles for gases

At the same temperature and pressure, the volume of a gas is directly proportional to the number of moles it contains (Avogadro's Law). Therefore, we can write the proportion: $$ \frac{V_{NH_3}}{V_{NO}} = \frac{n_{NH_3}}{n_{NO}} $$ where \(V_{NH_3}\) and \(V_{NO}\) are the volumes of Ammonia and Nitrogen oxide, and \(n_{NH_3}\) and \(n_{NO}\) are the moles of Ammonia and Nitrogen oxide, respectively. We are given that \(V_{NO} = 12.8 \mathrm{~L}\).

Use the stoichiometry from the balanced equation to relate the moles of Ammonia and Nitrogen oxide

From the stoichiometry of the balanced equation, we know that: $$ \frac{n_{NH_3}}{n_{NO}} = \frac{4}{6} $$

Solve for the volume of Ammonia

Now, we can substitute the relationship between moles in the formula for volumes: $$ \frac{V_{NH_3}}{12.8 \mathrm{~L}} = \frac{4}{6} $$ Solving for \(V_{NH_3}\), we get: $$ V_{NH_3} = 12.8 \mathrm{~L} \cdot \frac{4}{6} = 8.533 \mathrm{~L} $$ Thus, 8.533 L of Ammonia are required to change 12.8 L of Nitrogen oxide to Nitrogen gas.

Key concepts

Avogadro's Law

Avogadro's Law plays a fundamental role in understanding the behavior of gases during chemical reactions. Simply put, it states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. This is key in stoichiometry of gas reactions because it allows us to directly relate the volume of gas to the number of moles it contains.

When we say one mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters, we're using Avogadro's Law. For the given problem, even though we aren't dealing with STP conditions, the law still applies because the gases are measured at the same conditions of temperature and pressure. This means that if we have the volume of one gas, like nitrogen oxide, we can use the moles to volume ratio of the balanced chemical equation to find the volume of another reactant or product gas, such as ammonia.

Balanced Chemical Equations

Balanced chemical equations are the foundation of stoichiometry. They tell us the exact proportions of reactants and products involved in a chemical reaction, ensuring the conservation of atoms according to the Law of Conservation of Mass. The coefficients in a balanced equation represent the relative number of moles of each substance involved.

In our exercise, the balanced chemical equation shows that 6 moles of nitrogen oxide react with 4 moles of ammonia to form 5 moles of nitrogen gas and 6 moles of water. The equation balances out in terms of atoms on both sides, but it also provides us with the important mole ratio between reactants. This ratio (4 moles of NH₃ per 6 moles of NO) is crucial because it allows us to calculate the amount of ammonia needed to react with a given volume of nitrogen oxide.

Gas Volume-Mole Relationship

The gas volume-mole relationship is an application of Avogadro's Law and a vital part of solving stoichiometric problems involving gases. According to this relationship, the volume of a gas is directly proportional to the amount of substance (moles) when temperature and pressure are held constant. Therefore, we can write the proportion as \( \frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}} \), where \( V_{1} \) and \( V_{2} \) are volumes of two gases, and \( n_{1} \) and \( n_{2} \) are their respective moles.

Applying this to our problem, we can use the given volume of nitrogen oxide and the mole ratio from the balanced equation to find the required volume of ammonia. Since we have a direct relationship between the volume of nitrogen oxide and the volume of ammonia via their mole ratio, we can solve the problem without needing to know the actual number of moles present, providing an elegant and efficient solution to the stoichiometry problem.