Question

Large quantities of ammonia are burned in the presence of a platinum catalyst to give nitric oxide, as the first step in the preparation of nitric acid. \(\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \stackrel{\mathrm{Pt}}{\longrightarrow} \mathrm{NO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (Unbalanced) Suppose a vessel contains \(0.12\) moles \(\mathrm{NH}_{3}\) and \(0.14\) moles \(\mathrm{O}_{2}\). How many moles of NO may be obtained?

Short answer

A: 0.112 moles of NO can be obtained from the given amount of reactants.

Step-by-step solution

Balance the chemical equation

First, we need to balance the given chemical equation: \(\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\stackrel{\mathrm{Pt}}{\longrightarrow} \mathrm{NO}(\mathrm{g})+\mathrm{H}_{2}\mathrm{O}(\mathrm{g})\). Balancing the equation gives: \(4\mathrm{NH}_{3}(\mathrm{~g})+5\mathrm{O}_{2}(\mathrm{~g})\stackrel{\mathrm{Pt}}{\longrightarrow} 4\mathrm{NO}(\mathrm{g})+6\mathrm{H}_{2}\mathrm{O}(\mathrm{g})\)

Identify the limiting reactant

We now have the balanced equation, and we know the initial moles of \(\mathrm{NH}_{3}\) and \(\mathrm{O}_{2}\). To find the limiting reactant, we will compare the mole ratios of \(\mathrm{NH}_{3}\) to \(\mathrm{O}_{2}\) in the balanced equation with our given moles: Initial moles of \(\mathrm{NH}_{3} = 0.12\) moles Initial moles of \(\mathrm{O}_{2} = 0.14\) moles From the balanced equation, the mole ratio of \(\mathrm{NH}_{3}\) to \(\mathrm{O}_{2}\) is \(4:5\). We need to find which reactant is limiting by comparing the given amounts to the required ratios. Let's divide the given moles by their respective stoichiometric coefficients: \(\frac{0.12}{4} = 0.03\) \(\frac{0.14}{5} = 0.028\) The smallest value, 0.028, corresponds to the mole ratio of \(\mathrm{O}_{2}\), meaning that \(\mathrm{O}_{2}\) is the limiting reactant.

Calculate the moles of NO formed

Now that we've identified the limiting reactant, we can use the stoichiometric coefficients from the balanced equation to determine the moles of NO formed. Recall that the mole ratio of \(\mathrm{O}_{2}\) to \(\mathrm{NO}\) is \(5:4\). Since we started with 0.14 moles of \(\mathrm{O}_{2}\), we can calculate the moles of NO formed: \(\frac{4\,\text{moles of NO}}{5\,\text{moles of}\,\mathrm{O}_{2}}=\frac{x\,\text{moles of NO}}{0.14\,\text{moles of}\,\mathrm{O}_{2}}\) Cross-multiplying yields: \(5x = 0.14\times4\) Now, we solve for x: \(x = \frac{0.14\times4}{5} = 0.112\) So 0.112 moles of NO may be obtained from the given amount of reactants.

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental step in understanding chemical reactions. It ensues adherence to the Law of Conservation of Mass, which states that matter cannot be created or destroyed in an ordinary chemical reaction. Therefore, the same number of atoms of each element must appear on both the reactant and product sides of the equation.

What makes an equation balanced? It is one where the number of atoms for each element in the reaction and the total charge are the same on both sides. This is achieved by adjusting the coefficients (the numbers in front of the chemical formulas), not the subscripts in the formulas themselves. To illustrate, in the provided exercise, the balanced form of the ammonia oxidation reaction is:
\[4\mathrm{NH}_3(\mathrm{~g})+5\mathrm{O}_2(\mathrm{~g})\stackrel{\mathrm{Pt}}{\longrightarrow}4\mathrm{NO}(\mathrm{g})+6\mathrm{H}_2\mathrm{O}(\mathrm{g})\]
This balance is reached by ensuring that there are equal numbers of nitrogen, hydrogen, and oxygen atoms on both sides. Practicing this skill improves problem-solving abilities and paves the way for further analysis such as identifying the limiting reactant and calculating mole ratios.

Limiting Reactant

In chemical reactions, the limiting reactant (or limiting reagent) is the substance that is entirely consumed first and therefore determines the maximum amount of product that can be produced. This concept is analogous to the idea of a bottleneck in a manufacturing process. Once the limiting reactant is used up, no further reaction can occur, and any remaining reactants are considered excess.

To determine the limiting reactant, one must compare the mole ratios of the reactants to the coefficients in the balanced chemical equation. In the given exercise, upon comparison of the actual mole amounts:
\[\frac{0.12}{4} = 0.03\]
\[\frac{0.14}{5} = 0.028\]
The smaller ratio indicates which reactant will run out first. In this case, oxygen (\(\mathrm{O}_2\)) is the limiting reactant. This understanding allows for the accurate prediction of the amounts of products formed and is crucial when planning reactions for chemical synthesis in both academic and industrial settings.

Mole Ratio

The mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation. It allows for the conversion between moles of reactant and moles of product, as well as between moles of different reactants. This concept plays a critical role in stoichiometric calculations; wherein chemists use it to quantify relationships in a chemical reaction. Specifically, the mole ratio indicates the proportion in which substances react based on their molecular weights and Avogadro's number (6.022 x \(10^{23}\) particles per mole).

Using the mole ratio from the balanced equation in the sample exercise, and knowing the amount of the limiting reactant, we can calculate the expected product. In the following step:
\[\frac{4\text{ moles of NO}}{5\text{ moles of }\mathrm{O}_2}=\frac{x\text{ moles of NO}}{0.14\text{ moles of }\mathrm{O}_2}\]
The cross-multiplication used is standard in finding an unknown factor in direct proportion situations. As we solve for x, we find that 0.112 moles of nitric oxide (NO) can be formed. The mole ratio concept allows us to put this reaction into perspective, tying together the balanced equation and the limiting reactant findings to effectively quantify the expected output of a given chemical reaction.