Question

Consider the following reaction: \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) If a container were to have only 10 molecules of \(\mathrm{O}_{2}(g)\) and 10 molecules of \(\mathrm{NH}_{3}(g),\) how many total molecules (reactant and product) would be present in the container after the above reaction goes to completion? b. Using "microscopic" pictures, draw the total molecules present inside the container after the reaction occurs. c. What mass of \(\mathrm{NO}(g)\) is present in the container after the reaction occurs? (Report your final answer to 4 significant figures.

Short answer

After the reaction goes to completion, there will be \(22\) total molecules present in the container. The container will have \(8\) molecules of NO, \(12\) molecules of H₂O, and \(2\) unreacted NH₃ molecules. The mass of NO produced is \(3.987 \times 10^{-22} \mathrm{g}\), reported to 4 significant figures.

Step-by-step solution

Identify the limiting reactant

Compare the given ratio of reactants to the stoichiometric ratio: Given ratio: NH₃:O₂ = 10 molecules:10 molecules = 1:1 Stoichiometric ratio: NH₃:O₂ = 4 molecules:5 molecules When compared, we can see that there are more molecules of NH₃ than required for the given number of O₂ molecules. Therefore, O₂ will be the limiting reactant.

Calculate the number of molecules produced after the reaction

Use the stoichiometric coefficients to find the number of molecules produced for both NO and H₂O: For NO: 5 O₂ produce 4 NO, so 10 O₂ produce 10 * (4 NO / 5 O₂) = 8 NO molecules For H₂O: 5 O₂ produce 6 H₂O, so 10 O₂ produce 10 * (6 H₂O / 5 O₂) = 12 H₂O molecules

Calculate the number of NH₃ molecules remaining

Since O₂ is the limiting reactant, not all NH₃ molecules will be consumed. We can calculate the unreacted NH₃ molecules: Initial NH₃ molecules = 10 Consumed NH₃ molecules = 10 * (4 NH₃ / 5 O₂) = 8 Remaining NH₃ molecules = 10 - 8 = 2

Find the total number of molecules after the reaction

Add up the molecules present after the reaction completes, including reactants and products: Total molecules = Remaining NH₃ molecules + NO molecules + H₂O molecules Total molecules = 2 + 8 + 12 = 22 There will be 22 total molecules present in the container after the reaction goes to completion. b. Drawing the total molecules present inside the container

Draw NO molecules

Draw 8 nitrogen monoxide (NO) molecules as a single nitrogen (N) atom bonded to a single oxygen (O) atom.

Draw H₂O molecules

Draw 12 water (H₂O) molecules as a single oxygen (O) atom bonded to two hydrogen (H) atoms.

Draw NH₃ molecules

Draw 2 unreacted ammonia (NH₃) molecules as a nitrogen (N) atom bonded to three hydrogen (H) atoms. c. Finding the mass of NO produced

Calculate the number of moles of NO produced

We know that there are 8 NO molecules produced, so we can determine the number of moles of NO using Avogadro's number: Moles of NO = 8 molecules / (6.022 x 10²³ molecules/mol) ≈ 1.329 x 10⁻²³ mol

Calculate the mass of NO produced

Determine the molar mass of NO (14 g/mol for N and 16 g/mol for O gives a total of 30 g/mol). Then, multiply the moles of NO by the molar mass to find the mass of NO produced: Mass of NO = (1.329 x 10⁻²³ mol) x (30 g/mol) ≈ 3.987 x 10⁻²² g

Report the final answer to 4 significant figures

The mass of NO produced after the reaction is 3.987 x 10⁻²² g, reported to 4 significant figures.

Key concepts

Limiting Reactant

In chemistry, the concept of a limiting reactant is crucial when deciphering the outcome of a chemical reaction. It refers to the substance that is entirely consumed first, dictating the extent to which a reaction can proceed before stopping. As in the exercise, to identify the limiting reactant, we compare the ratio of the reactant's molecules present with the ratio required by the balanced chemical reaction, or the stoichiometric ratios.

The reaction given in our example is balanced and provides the stoichiometric coefficients needed to determine the limiting reactant. It can often be a challenge for students to visualize why one reactant limits the reaction. A helpful way to imagine this scenario is to think of it as a simple recipe. If a recipe calls for four eggs and five cups of flour to make a batch of cookies, but you only have four eggs and can't find the fifth, you won't be able to complete another batch, regardless of how much flour you have left. In this case, eggs would be the limiting reactant.

The process of finding the limiting reactant involves a comparison between the actual mole ratio of reactants and the stoichiometric mole ratio from the balanced equation. Once we establish which reactant will run out first, we can predict the amounts of products formed and any excess reactants remaining.

Stoichiometric Coefficients

Stoichiometric coefficients are the numbers in front of molecules in a balanced chemical equation and reflect the proportional amounts of reactants and products involved in a reaction. They serve as a guide to the 'recipe' of a chemical reaction, telling us how many particles of each substance are needed or produced.

For example, the coefficients in our exercise's reaction equation tell us that 4 molecules of ammonium (\( \text{NH}_3 \text{(g)} \)) react with 5 molecules of oxygen (\( \text{O}_2 \text{(g)} \) ) to produce 4 molecules of nitrogen monoxide (\( \text{NO} \text{(g)} \) ) and 6 molecules of water vapor (\( \text{H}_2 \text{O} \text{(g)} \) ). These coefficients are the key to any stoichiometry calculation. When we understand that these numbers represent molar ratios, not just molecule-to-molecule ratios, we unlock the ability to scale reactions up to observable amounts like grams or mol.

Understanding these coefficients is fundamental for students, as they can then use this knowledge to calculate the amounts of products formed and reactants consumed when given any one quantity.

Molecular Representation of Reactions

The molecular representation of reactions involves illustrating substances at the molecular level to depict the changes that occur during a chemical reaction. This illustration can be a powerful tool for visual learners to grasp complex concepts. In the exercise solution, we draw the reactants and products based on their chemical formulas and the stoichiometric coefficients provided.

Molecular visuals help elucidate details like the orientation of atoms and the ratios in which they combine or break apart. Seeing a graphical representation of, for example, 8 NO molecules and 12 H2O molecules produced from 10 O2 and 10 NH3 molecules gives students an immediate visual impression of the reaction's results, including the two remaining NH3 molecules that were not used up.

For increased clarity, these images are sometimes accompanied by microscopic pictures or molecular models in textbooks and resource materials. This not only serves to make the concept more tangible but also aids students in associating the abstract symbols and formulas with actual substances.

Molar Mass Calculation

Molar mass calculation involves determining the mass of one mole of a substance. It is a fundamental aspect of stoichiometry, as it allows us to convert between moles and grams - a necessary step for connecting the molecular scale to the macroscopic world that we can measure.

In the given exercise, understanding the molar mass calculation was applied in the final step for finding the mass of nitrogen monoxide (NO) produced. The molar mass is calculated by summing the atomic masses of each atom in a compound. For instance, NO has one nitrogen atom with an atomic mass of approximately 14 grams per mole (g/mol) and one oxygen atom with an atomic mass of 16 g/mol, resulting in a molar mass of 30 g/mol for NO.

These calculations are not only used in laboratory settings for measuring substances but also for balancing chemical equations and converting between grams, moles, and the number of molecules - essential tasks in many chemical reactions. Calculating molar mass accurately is an indispensable skill for students to master in order to perform well in chemistry.

Avogadro's Number

Avogadro's number is a constant that represents the amount of entities, usually atoms or molecules, in one mole of a substance. Its value is approximately \(6.022 \times 10^{23}\) entities per mole and is a bridge between the atomic scale and the real-world scale. In our exercise, Avogadro's number is used to convert from an actual number of molecules to moles, a step involved in the calculation of the mass of a substance.

Students often struggle with comprehending the magnitude of Avogadro's number, but it can be compared to understanding vast distances in astronomy or the tiny scales of particles. Highlighting its use in various chemical calculations can solidify its importance - whether for computing the mass of products formed in a reaction, as shown in our example, determining the number of particles in a given quantity of material, or in other applications where mole-to-molecule conversions are needed.

It's important for students to practice with Avogadro's number to build a strong foundation in their comprehension of chemistry, as it is a key part of the quantitative descriptions that are essential to the science.