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.)

Step by step solution

01

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

02

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: $$\frac{5}{4} \cdot x + \frac{7}{4} \cdot y$$

03

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 \(\mathrm{O}_{2}(g)\): $$10.00 - \frac{5}{4}\cdot x - \frac{7}{4}\cdot y = 6.75$$

04

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: $$x = 2.00 - 0.50 = 1.50 \mathrm{~mol}$$

05

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.

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