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 \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, 6.75 moles of \(\mathrm{O}_{2}(g)\) 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 reaction equations

Let's balance the two given unbalanced reactions. Reaction 1: \[\mathrm{4NH}_{3}(g)+\mathrm{5O}_{2}(g) \longrightarrow 4\mathrm{NO}(g)+6\mathrm{H}_{2}\mathrm{O}(g)\] Reaction 2: \[\mathrm{4NH}_{3}(g)+\mathrm{7O}_{2}(g) \longrightarrow 4\mathrm{NO}_{2}(g)+6\mathrm{H}_{2} \mathrm{O}(g)\]

02

Determine moles of \(\mathrm{O}_{2}\) consumed

Since we know that there are 10.00 moles of \(\mathrm{O}_{2}\) initially and 6.75 moles of \(\mathrm{O}_{2}\) left after the reaction, the moles of \(\mathrm{O}_{2}\) consumed can be calculated as follows: Moles of \(\mathrm{O}_{2}\) consumed = Initial moles of \(\mathrm{O}_{2}\) - Remaining moles of \(\mathrm{O}_{2}\) Moles of \(\mathrm{O}_{2}\) consumed = \(10.00 - 6.75 = 3.25\) moles

03

Calculate the moles of each product using the balanced reactions

In Reaction 1, we can notice that for every 5 moles of \(\mathrm{O}_{2}\) consumed, 4 moles of \(\mathrm{NO}\) are formed. In Reaction 2, we can notice that for every 7 moles of \(\mathrm{O}_{2}\) consumed, 4 moles of \(\mathrm{NO}_{2}\) are formed. Now, let's consider that 'a' moles of \(\mathrm{NO}\) are produced via Reaction 1, and 'b' moles of \(\mathrm{NO}_{2}\) are produced via Reaction 2. Since only 3.25 moles of \(\mathrm{O}_{2}\) are consumed in total, we can set up the following equation using the stoichiometric ratios in the balanced reactions: \[\frac{a}{5} + \frac{b}{7} = 3.25\]

04

Use moles of \(\mathrm{NH}_{3}\) to find 'a'

Since all \(\mathrm{NH}_{3}\) is consumed in the reactions, we can use its moles to find the value of 'a'. There are 2.00 moles of \(\mathrm{NH}_{3}\) initially. From Reaction 1, for every 4 moles of \(\mathrm{NH}_{3}\) consumed, 4 moles of \(\mathrm{NO}\) are produced. Therefore, \[\frac{a}{4} = 2\] Solve for 'a': \[a = 4\]

05

Find 'b' using 'a' value

Substitute the value of 'a' found in the previous step back into the equation for moles of \(\mathrm{O}_{2}\) consumed: \[\frac{4}{5} + \frac{b}{7} = 3.25\] Now, solve for 'b': \[\frac{b}{7} = 3.25 - \frac{4}{5}\] \[\frac{b}{7} = \frac{16.25 - 4}{5} = \frac{12.25}{5}\] \[b = \frac{85.75}{5}\] Since 'b' represents moles of \(\mathrm{NO}_{2}\) produced and we're only interested in the moles of \(\mathrm{NO}\) produced, the value of 'a' provides our final answer.

06

Final answer

The number of moles of \(\mathrm{NO}\) in the product mixture: \(a = 4\) moles.

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