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Chapter 4: Problem 63

How many moles of \(\mathrm{NO}(\mathrm{g})\) can be produced in the reaction of \(3.00 \mathrm{mol} \mathrm{NH}_{3}(\mathrm{g})\) and \(4.00 \mathrm{mol} \mathrm{O}_{2}(\mathrm{g}) ?\) $$ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

Short Answer

Expert verified
The maximum possible amount of \(\mathrm{NO}\) that can be produced is 3.00 moles.

Step by step solution

01

Analyzing the Chemical Reaction

Look at the balanced chemical equation: \(4 \mathrm{NH}_{3} + 5 \mathrm{O}_{2} \longrightarrow 4 \mathrm{NO} + 6 \mathrm{H}_{2}\mathrm{O} \). This tells us the molar relationship between reactants and products. The coefficients represent the number of moles. In this case for every 4 moles of \(\mathrm{NH}_{3}\) and 5 moles of \(\mathrm{O}_{2}\), 4 moles of \(\mathrm{NO}\) and 6 moles of \(\mathrm{H}_{2}\mathrm{O}\) are produced.
02

Identify the Limiting Reagent

A reaction goes to completion when one of the reactants gets completely consumed, which is referred to as the limiting reagent. We have: 1. For 4 moles of \(\mathrm{NH}_{3}\), we need \(5/4 \times 4 = 5\) moles of \(\mathrm{O}_{2}\). But we have 4.00 moles of \(\mathrm{O}_{2}\), 2. For 5 moles of \(\mathrm{O}_{2}\), we need \(4/5 \times 5 = 4\) moles of \(\mathrm{NH}_{3}\). We have 3.00 moles of \(\mathrm{NH}_{3}\).So, \(\mathrm{NH}_{3}\) is the limiting reagent.
03

Calculate the Amount of Product

The amount of product formed is proportional to the amount of the limiting reagent. Since 4 moles of \(\mathrm{NH}_{3}\) produces 4 moles of \(\mathrm{NO}\), 3.00 moles of \(\mathrm{NH}_{3}\) (the limiting reagent) will produce \(4/4 \times 3 = 3.00\) moles of \(\mathrm{NO}\). Therefore, the maximum amount of \(\mathrm{NO}\) that can be produced is 3.00 moles.

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