Question

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(g)\) or \(\mathrm{NO}_{2}(g)\) according to these unbalanced equations: $$\begin{array}{l}\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\\\\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\end{array}$$ In a certain experiment 2.00 moles of \(\mathrm{NH}_{3}(g)\) and 10.00 moles of \(\mathbf{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, 6.75 moles of \(\mathbf{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

Short answer

In the product mixture, there are \(0.752\) moles of \(\mathrm{NO}(g)\) formed.

Step-by-step solution

Balancing the reactions

First, we need to balance the given reactions: 1. \(\mathrm{NH}_{3}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g) + \mathrm{H}_{2} \mathrm{O}(g)\) 4\(\mathrm{NH}_{3}(g) + 5\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{NO}(g) + 6\mathrm{H}_{2} \mathrm{O}(g)\) 2. \(\mathrm{NH}_{3}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(g)\) 4\(\mathrm{NH}_{3}(g) + 7\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{NO}_{2}(g) + 6\mathrm{H}_{2} \mathrm{O}(g)\)

Determine the moles of \(\mathrm{O}_{2}\) used in each reaction

We know that 10.00 moles of \(\mathrm{O}_{2}(g)\) were initially present, and after the reaction, 6.75 moles remain. This means that: Total moles of \(\mathrm{O}_{2}\) used = Initial moles of \(\mathrm{O}_{2}\) - Remaining moles of \(\mathrm{O}_{2}\) Total moles of \(\mathrm{O}_{2}\) used = 10.00 moles - 6.75 moles = 3.25 moles Now, let's consider that \(y\) moles of \(\mathrm{O}_{2}(g)\) are used in the first reaction and \(x\) moles of \(\mathrm{O}_{2}(g)\) are used in the second reaction. Then, we can write the following equation: \(x + y = 3.25\)

Calculate the moles of \(\mathrm{NO}(g)\) formed in the reaction

In the balanced reaction 1, 4 moles of \(\mathrm{NH}_{3}(g)\) react with 5 moles of \(\mathrm{O}_{2}(g)\) to produce 4 moles of \(\mathrm{NO}(g)\). From this information, we can get a proportionality constant, which we can use to calculate the moles of \(\mathrm{NO}(g)\) formed when a specific amount of \(\mathrm{O}_{2}(g)\) is consumed in the reaction. In reaction 1, we have: \(\frac{\text{Moles of \)\mathrm{NO}\( formed}}{\text{Moles of \)\mathrm{O}_{2}\( consumed}} = \frac{4}{5}\) Now, we know that 2.00 moles of \(\mathrm{NH}_{3}(g)\) is completely consumed in both reactions. We can write a similar relation of moles for reaction 2: \(\frac{\text{Moles of \)\mathrm{NH}_{3}\( consumed in reaction 1}}{\text{Moles of \)\mathrm{NH}_{3}\( consumed in reaction 2}} = \frac{4}{4}\) \(\frac{2 - y/5}{x/7} = 1\) Now we have two equations: 1. \(x + y = 3.25\) 2. \(2 - y/5 = x/7\) By solving these two equations simultaneously for \(x\) and \(y\), we get: \(x = 2.31\) (moles of \(\mathrm{O}_2\) consumed in the second reaction) \(y = 0.94\) (moles of \(\mathrm{O}_2\) consumed in the first reaction) Now we can find the moles of \(\mathrm{NO}(g)\) formed in the first reaction: Moles of \(\mathrm{NO}(g)\) = \(\frac{\text{4 moles}}{\text{5 moles of \)\mathrm{O}_{2}\( consumed}} \times \text{0.94 moles of \)\mathrm{O}_{2}\( consumed}\) Moles of \(\mathrm{NO}(g)\) = \(\frac{4}{5} \times 0.94\) Moles of \(\mathrm{NO}(g)\) formed = \(\boxed{0.752}\)

Key concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. It allows chemists to predict how much of a substance will be consumed or produced in a given reaction. Understanding stoichiometry involves using the molar ratios found in balanced chemical equations. These ratios are essential to calculating the exact amounts of chemicals needed or produced.

For example, in our exercise, we determine how many moles of ammonia (\(\text{NH}_3\)) react with oxygen (\(\text{O}_2\)) to form nitrogen monoxide (\(\text{NO}\)) or nitrogen dioxide (\(\text{NO}_2\)). By utilizing stoichiometry, we can calculate that 3.25 moles of \(\text{O}_2\) were used in the reaction, helping us find how much \(\text{NO}\) was produced. This concept empowers us to precisely predict and control chemical processes, making stoichiometry an invaluable tool in chemistry.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They showcase the substances involved, both reactants and products, and their relative quantities. To accurately convey these transformations, chemical equations must be balanced. Balancing ensures that the number of atoms of each element is the same on both sides of the equation, adhering to the Law of Conservation of Mass.

In the given problem, ammonia reacts with oxygen to create nitrogen oxides and water. The equations provided initially were unbalanced:

• \(\text{NH}_3(g) + \text{O}_2(g) \longrightarrow \text{NO}(g) + \text{H}_2\text{O}(g)\)
• \(\text{NH}_3(g) + \text{O}_2(g) \longrightarrow \text{NO}_2(g) + \text{H}_2\text{O}(g)\)

By balancing these, we ensure that chemical identities and mass are conserved. Balancing helps us derive the correct stoichiometric relationships necessary for calculations.

Reaction Balancing

Reaction balancing is the process of equalizing the number of atoms for each element on both sides of a chemical equation. It’s a fundamental requirement in chemistry to maintain the law of conservation of mass. The balanced form of a chemical equation gives insight into the stoichiometric relationships between reactants and products.

In our problem, the reactions were balanced as follows:

• \(4\text{NH}_3(g) + 5\text{O}_2(g) \longrightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g)\)
• \(4\text{NH}_3(g) + 7\text{O}_2(g) \longrightarrow 4\text{NO}_2(g) + 6\text{H}_2\text{O}(g)\)

Balancing these equations allowed us to apply the stoichiometric ratios effectively. For instance, knowing that 4 moles of \(\text{NH}_3\) yield 4 moles of \(\text{NO}\) while consuming 5 moles of \(\text{O}_2\), we can correctly deduce the amount of \(\text{NO}\) formed, as demonstrated in the exercise where we calculated 0.752 moles of \(\text{NO}\). Reaction balancing transforms a theoretical equation into a practical framework for conducting and understanding chemical reactions.