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.)

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\(\mathrm{NH}_{3}(g) + 7\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{NO}_{2}(g) + 6\mathrm{H}_{2} \mathrm{O}(g)\)

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: \(x + y = 3.25\)

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 = \(\boxed{0.752}\)