Problem 92
Question
Nitrogen reacts with hydrogen to form ammonia. Represent each nitrogen atom by a square and each hydrogen atom with a circle. Starting with five molecules of both hydrogen and nitrogen, show pictorially what you have after the reaction is complete.
Step-by-Step Solution
Verified Answer
Answer: After the reaction, the final pictorial representation is:
Unreacted nitrogen (N_2) molecules: □ □ □ □
Unreacted hydrogen (H_2) molecules: ○ ○
Produced ammonia (NH_3) molecules: NH_3 NH_3
1Step 1: Understand the chemical reaction
The chemical reaction between nitrogen and hydrogen to form ammonia can be represented as follows:
N_2 + 3H_2 -> 2NH_3
Here, one molecule of nitrogen (N_2) reacts with three molecules of hydrogen (H_2) to form two molecules of ammonia (NH_3).
2Step 2: Represent given nitrogen and hydrogen molecules
We start with five molecules of both hydrogen (circles) and nitrogen (squares). Since nitrogen exists as a diatomic molecule (N_2), we will have 5 nitrogen molecules (10 nitrogen atoms), and 5 hydrogen molecules (10 hydrogen atoms).
Nitrogen molecules (N_2) : □ □ □ □ □
Hydrogen molecules (H_2) : ○ ○ ○ ○ ○
3Step 3: Perform the reaction
Now, we will let the nitrogen and hydrogen molecules react to form ammonia following the reaction previously described. We will use up as many nitrogen and hydrogen molecules as possible based on the stoichiometry of the reaction:
1 N_2 molecule + 3 H_2 molecules -> 2 NH_3 molecules
Since we have 5 nitrogen and 5 hydrogen molecules, it's crucial to notice that our limiting reactant is hydrogen in this case. We have enough hydrogen to react with 5/3 ≈ 1.67 nitrogen molecules. Therefore, only one nitrogen molecule can fully react with three hydrogen molecules to form ammonia.
Reaction outcome:
1 N_2 molecule + 3 H_2 molecules -> 2 NH_3 molecules
□ + ○ ○ ○ -> NH_3 NH_3
4Step 4: Show the final pictorial representation
Considering one nitrogen molecule has reacted with three hydrogen molecules to form two ammonia molecules, this is the pictorial representation of the system after the reaction is complete:
Unreacted nitrogen (N_2) molecules: □ □ □ □
Unreacted hydrogen (H_2) molecules: ○ ○
Produced ammonia (NH_3) molecules: NH_3 NH_3
Key Concepts
Chemical Reaction StoichiometryLimiting ReactantDiatomic Molecules
Chemical Reaction Stoichiometry
Chemical reaction stoichiometry is a fundamental aspect of chemistry that involves understanding the quantitative relationships between reactants and products in a chemical reaction. In the reaction where nitrogen and hydrogen form ammonia, the balanced chemical equation is \(N_2 + 3H_2 \rightarrow 2NH_3\). This equation shows that one molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia. To balance a chemical reaction, it is essential to ensure that the number of atoms of each element is the same on both sides of the equation. Here, the nitrogen and hydrogen atoms are balanced. This simple but powerful concept provides insight into how much of each reactant is required and how much product can be made. Understanding stoichiometry enables chemists to calculate the quantities of substances consumed and produced, letting them plan and control chemical processes effectively. The idea doesn't just stop at balancing equations; it also involves conversions such as moles and mass ratio calculations. For example, knowing that one mole of \(N_2\) reacts with three moles of \(H_2\) to produce two moles of \(NH_3\) allows chemists to scale reactions, whether they need to create a small amount or an industrial-scale batch.
Limiting Reactant
In chemical reactions, the limiting reactant is the substance that is entirely consumed first, stopping the progress of the reaction. It determines the maximum amount of product that can be formed. In our ammonia synthesis exercise, we start with five molecules each of nitrogen and hydrogen. The balanced equation, \(N_2 + 3H_2 \rightarrow 2NH_3\), indicates that three molecules of hydrogen are required for every molecule of nitrogen to form ammonia. With only five hydrogen molecules present, hydrogen quickly becomes the limiting reactant. We can react a maximum of one nitrogen molecule with three hydrogen molecules to produce two ammonia molecules because after this, hydrogen runs out, even though we have excess nitrogen left.Identifying the limiting reactant is crucial in both theoretical calculations and practical applications. It helps in optimizing the use of reactants and understanding maximum possible yields, saving time and resources in laboratory and industrial processes.
Diatomic Molecules
Diatomic molecules are a specific type of molecule composed of only two atoms, which can be either of the same element or different. In the ammonia synthesis reaction, nitrogen (N_2) and hydrogen (H_2) are both diatomic molecules. This means each molecule consists of two atoms bonded together.Understanding diatomic molecules like \(N_2\) and \(H_2\) is essential because they reflect how elements behave in their natural states. Most gases, including oxygen \(O_2\), chlorine \(Cl_2\), as well as hydrogen and nitrogen, exist naturally as diatomic molecules. This property is crucial in reaction equations, as the diatomic nature impacts the stoichiometry. For instance, when calculating amounts needed for reaction, one must consider these elements as \(N_2\) or \(H_2\) rather than individual atoms. This ensures that balanced equations reflect real-world conditions more accurately and aids in the precise measurement of reactants and products. Understanding the diatomic nature of these molecules thus forms a foundational component of molecular chemistry.
Other exercises in this chapter
Problem 88
Perhaps the simplest way to calculate Avogadro's number is to compare the charge on the electron, first determined by Robert Millikan in 1909, with the charge o
View solution Problem 89
Given a pair of elements and their mass relation, answer the following questions. (a) The mass of 4 atoms of \(A=\) the mass of 6 atoms of \(B\). Which element
View solution Problem 94
When \(4.0 \mathrm{~mol}\) of \(\mathrm{CCl}_{4}\) reacts with an excess of \(\mathrm{HF}, 3.0 \mathrm{~mol}\) of \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) (Freon) i
View solution Problem 95
Suppose that the atomic mass of \(\mathrm{C}-12\) is taken to be \(5.000 \mathrm{amu}\) and that a mole is defined as the number of atoms in \(5.000 \mathrm{~kg
View solution