Question

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(\mathrm{g})\) or \(\mathrm{NO}_{2}(\mathrm{~g})\) according to these unbalanced equations: $$ \begin{array}{l} \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ In a certain experiment \(2.00 \mathrm{~mol} \mathrm{NH}_{3}(g)\) and \(10.00 \mathrm{~mol}\) \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, \(6.75 \mathrm{~mol} \mathrm{O}_{2}(g)\) 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

The number of moles of \(\mathrm{NO}(g)\) in the product mixture is 0.375 mol.

Step-by-step solution

Balance the given chemical reactions.

To balance the given reactions, we should make sure there are equal numbers of each element on both sides of the reaction. Balanced equation 1: $$4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ Balanced equation 2: $$4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$

Express the moles of products in terms of moles of consumed ammonia.

We'll use the symbols \(x\) and \(y\) for moles of consumed ammonia in reaction 1 and reaction 2, respectively. Moles of \(\mathrm{NO}(g)\) produced in reaction 1: $$\frac{1}{4} \cdot x$$ Moles of \(\mathrm{NO}_{2}(g)\) produced in reaction 2: $$\frac{1}{4} \cdot y$$ Moles of \(\mathrm{H}_{2}\mathrm{O}(g)\) produced in both reactions: $$\frac{3}{2} (x + y)$$ Moles of \(\mathrm{O}_{2}(g)\) consumed in both reactions: $$\frac{5}{4} \cdot x + \frac{7}{4} \cdot y$$

Set up a system of linear equations using the given constraints.

Based on the given information, we can set up the following system of linear equations: 1. Total consumed \(\mathrm{NH}_{3}(g)\): $$x + y = 2.00$$ 2. Total remaining \(\mathrm{O}_{2}(g)\): $$10.00 - \frac{5}{4}\cdot x - \frac{7}{4}\cdot y = 6.75$$

Solve the linear system.

To find the values of \(x\) and \(y\), we can solve this system of linear equations. First, we'll solve the first equation for \(x\) and substitute it in the second equation. Substituting the first equation into the second equation: $$10.00 - \frac{5}{4}(2.00 - y) - \frac{7}{4}\cdot y = 6.75$$ Solving for \(y\) gives us: $$y = 0.50 \mathrm{~mol}$$ Now we can find \(x\) using the first equation: $$x = 2.00 - 0.50 = 1.50 \mathrm{~mol}$$

Calculate the number of moles of \(\mathrm{NO}(g)\).

Finally, we'll use the value of \(x\) (moles of consumed \(\mathrm{NH}_{3}\) in reaction 1) to calculate the moles of \(\mathrm{NO}(g)\) produced: Moles of \(\mathrm{NO}(g)\): $$\frac{1}{4}\cdot x = \frac{1}{4} \cdot 1.50 = 0.375 \mathrm{~mol}$$ Hence, there are 0.375 mol of \(\mathrm{NO}(g)\) in the product mixture.

Key concepts

Stoichiometry

Stoichiometry is an essential concept in chemistry that involves the calculation of reactants and products in a chemical reaction. It is all about the quantitative relationships between different substances as they participate in chemical reactions. To master stoichiometry, it's crucial to understand the balanced chemical equation that represents these reactions. Based on the coefficients in the balanced equation, we can convert moles of different substances and determine their proportions. This tells us how much of one substance will react with another, or what quantity of product will result from given amounts of reactants.

Balancing Chemical Equations

Balancing chemical equations is a foundational task in chemistry that ensures the conservation of mass in a reaction. Each side of the equation must have the same number of atoms for each element, reflecting that atoms are neither created nor destroyed. For the ammonia reactions given, the initial equations need to be balanced so the number of atoms of nitrogen, hydrogen, and oxygen are equal on both sides:

- For the formation of \( ext{NO}\) and water, the balanced equation is: \[4 ext{NH}_3(g) + 5 ext{O}_2(g) ightarrow 4 ext{NO}(g) + 6 ext{H}_2 ext{O}(g)\]
- The reaction yielding \( ext{NO}_2\) requires: \[4 ext{NH}_3(g) + 7 ext{O}_2(g) ightarrow 4 ext{NO}_2(g) + 6 ext{H}_2 ext{O}(g)\]

These balanced equations serve as the foundation for stoichiometric calculations that describe how reactants are converted into products.

Ammonia Reaction

Ammonia, \( ext{NH}_3\), is a common reactant in industrial chemical processes. It reacts with oxygen gas \( ext{O}_2\) in two potential pathways—forming either nitrogen monoxide (\( ext{NO}\)) or nitrogen dioxide (\( ext{NO}_2\)). These reactions are crucial in various applications such as the manufacturing of fertilizers and the catalytic reduction of nitrogen oxides in pollution control.

Understanding these reactions requires recognition of the limiting reactants, the conservation laws, and the potential products formed. Through stoichiometry and balanced equations, chemists can delineate which reactions predominate under specific conditions and calculate the resultant amounts of products formed.

Gas Laws

Gas laws, such as Boyle's Law, Charles's Law, and Avogadro's Law, describe how gases behave under varying conditions of pressure, volume, and temperature. These laws are vital in interpreting the behavior of gaseous reactants and products in chemical reactions.

For reactions involving gases, such as the conversion of ammonia with oxygen, we might use gas laws to relate quantities of gases to each other. For instance, if the volume and pressure are held constant, the reaction's progress can be assessed by analyzing changes in the quantity (moles) of gases. Similarly, recognizing the number of moles allows us to understand the gaseous products formed and predict pressures and volumes during the reaction.