Problem 84 - Chemistry and Chemical Reactivity

Question

The reaction of \(750 .\) g each of \(\mathrm{NH}_{3}\) and \(\mathrm{O}_{2}\) was found to produce \(562 \text { g of } \mathrm{NO} \text { (see pages } 177-179)\). $$4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\ell)$$ (a) What mass of water is produced by this reaction? (b) What mass of \(\mathrm{O}_{2}\) is required to consume \(750 . \mathrm{g}\) of \(\mathrm{NH}_{3} ?\)

Step-by-Step Solution

Answer

(a) 506.34 g of water is produced. (b) 1764.8 g of \(\mathrm{O}_2\) is required.

Step 1: Calculate Moles of NH3 and O2

First find the number of moles of \(\mathrm{NH}_3\) and \(\mathrm{O}_2\) using their molar masses. The molar mass of \(\mathrm{NH}_3\) is approximately 17 \(\mathrm{g/mol}\), and for \(\mathrm{O}_2\), it is 32 \(\mathrm{g/mol}\). Calculate moles for both substances: \[\text{Moles of } \mathrm{NH}_3 = \frac{750}{17} \approx 44.12 \, \text{moles}, \quad \text{Moles of } \mathrm{O}_2 = \frac{750}{32} \approx 23.44 \, \text{moles}.\]

Step 2: Determine Limiting Reactant

By comparing the stoichiometric coefficients, 4 moles of \(\mathrm{NH}_3\) require 5 moles of \(\mathrm{O}_2\). Calculate how much \(\mathrm{O}_2\) is needed: \[ \text{Needed } \, \mathrm{O}_2 = \frac{5}{4} \times 44.12 \approx 55.15 \, \text{moles}. \] As 23.44 moles of \(\mathrm{O}_2\) are available, \(\mathrm{O}_2\) is the limiting reactant.

Step 3: Calculate Water Produced

Since \(\mathrm{O}_2\) is the limiting reactant, calculate the moles of \(\mathrm{H}_2\mathrm{O}\) produced from the moles of \(\mathrm{O}_2\): \[6 \text{ moles of } \mathrm{H}_2\mathrm{O} \text{ for 5 moles of } \mathrm{O}_2. \] Calculate moles of \(\mathrm{H}_2\mathrm{O}\): \[ \text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{6}{5} \times 23.44 \approx 28.13 \, \text{moles}. \] Convert moles of \(\mathrm{H}_2\mathrm{O}\) to mass using its molar mass (18 \(\mathrm{g/mol}\)):  \[ \text{Mass of } \mathrm{H}_2\mathrm{O} = 28.13 \times 18 \approx 506.34 \, \mathrm{g}. \]

Step 4: Calculate Required O2 for 750 g NH3

If \(\mathrm{NH}_3\) were limiting, find required \(\mathrm{O}_2\) for 44.12 moles of \(\mathrm{NH}_3\): \[ \text{Needed } \mathrm{O}_2 = \frac{5}{4} \times 44.12 \approx 55.15 \, \text{moles}. \] Thus, the mass of \(\mathrm{O}_2\) required is \[ \text{Mass of } \mathrm{O}_2 = 55.15 \times 32 \approx 1764.8 \, \mathrm{g}. \]

Key Concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that determines the amount of product formed. It is completely consumed in the reaction, and once it's used up, the reaction stops. Understanding which reactant is limiting helps you predict the quantities of products and leftovers accurately.

To identify the limiting reactant, compare the molar ratio of the reactants used in the reaction to the ratio described by the balanced chemical equation. For example, in the reaction between ammonia (\(\mathrm{NH}_3\)) and oxygen (\(\mathrm{O}_2\)) to form nitrogen monoxide (\(\mathrm{NO}\)), we find that 4 moles of \(\mathrm{NH}_3\) require 5 moles of \(\mathrm{O}_2\). Given that there are only 23.44 moles of \(\mathrm{O}_2\) compared to 44.12 moles of \(\mathrm{NH}_3\), oxygen becomes the limiting reactant, as we need more moles of it than what are present. This limitation dictates the maximum amount of \(\mathrm{H}_2\mathrm{O}\) produced in the reaction.

Molar Mass

Molar mass is a fundamental concept in stoichiometry, guiding you in converting between mass and moles, which are crucial for chemical calculations. Molar mass is defined as the mass of one mole of a substance, typically expressed in grams per mole (\(\mathrm{g/mol}\)).

For instance, ammonia (\(\mathrm{NH}_3\)) has a molar mass calculated by summing the atomic masses of nitrogen (approximately 14 \(\mathrm{g/mol}\)) and hydrogen (approximately 1 \(\mathrm{g/mol}\) per atom, with three atoms totaling 3 \(\mathrm{g/mol}\)), giving a total of approximately 17 \(\mathrm{g/mol}\). Similarly, the molar mass of oxygen (\(\mathrm{O}_2\)) is about 32 \(\mathrm{g/mol}\). These values allow you to determine how many moles are present in a given mass of a substance and are critical for calculating how much reactant is needed or product is formed in a reaction.

Chemical Reactions

Chemical reactions involve the transformation of reactants into one or more products. Each reaction is represented by a balanced chemical equation that shows the reactants and products along with their molar ratios. This helps in understanding the proportion of each substance involved.

For example, the reaction of ammonia (\(\mathrm{NH}_3\)) with oxygen (\(\mathrm{O}_2\)) is represented as:

• \(4\mathrm{NH}_3(\mathrm{g}) + 5\mathrm{O}_2(\mathrm{g}) \rightarrow 4\mathrm{NO}(\mathrm{g}) + 6\mathrm{H}_2\mathrm{O}(\ell)\)

This balanced equation indicates that for every 4 moles of \(\mathrm{NH}_3\), 5 moles of \(\mathrm{O}_2\) are needed, producing 4 moles of \(\mathrm{NO}\) and 6 moles of water (\(\mathrm{H}_2\mathrm{O}\)). The stoichiometry provides a clear guide of what quantities will react, aligning with the conservation of mass principle, as the number of each type of atom is the same on both sides of the equation.

Gaseous Reactions

Gaseous reactions involve reactants and/or products in the gas phase. These reactions are significant in various fields, including industry and environmental science. Understanding gaseous reactions allows you to manipulate conditions such as pressure, volume, and temperature, which directly influence reaction rates and product yields.

In the reaction between ammonia (\(\mathrm{NH}_3\)) and oxygen (\(\mathrm{O}_2\)) to form nitrogen monoxide (\(\mathrm{NO}\)) and water (\(\mathrm{H}_2\mathrm{O}\)), all reactants and one of the products are in the gaseous state. This reaction's behavior can be analyzed using the ideal gas law, \(PV = nRT\), where adjustments in pressure or volume impact reactant contribution and product formation. In real-world applications, controlling gaseous conditions can optimize reactions, making them more efficient and economically beneficial. With gases, it's essential to understand their nature since they occupy volume and are affected by temperature and pressure changes, distinguishing them from solids and liquids.