Problem 65 - Introductory Chemistry Atoms First

Question

The following chemical equation is unbalanced: \(\mathrm{I}_{2}+\mathrm{Cl}_{2} \rightarrow \mathrm{ICl}_{3}\) (a) Balance the equation. (b) The balanced equation is a source of many conversion factors containing the two reactants and one product. Write all of them that use the word mole(s). (c) The balanced equation is also a source of conversion factors involving the individual atoms and molecules. Write all of them that use the word mole(s).

Step-by-Step Solution

Answer

(a) The balanced chemical equation is \(\mathrm{I}_{2} + 3\mathrm{Cl}_{2} \rightarrow 2\mathrm{ICl}_{3}\). (b) Mole-based conversion factors: 1. \(\frac{1 \text{ mole of } \mathrm{I}_2}{3 \text{ moles of } \mathrm{Cl}_2}\) 2. \(\frac{3 \text{ moles of } \mathrm{Cl}_2}{1 \text{ mole of } \mathrm{I}_2}\) 3. \(\frac{1 \text{ mole of } \mathrm{I}_2}{2 \text{ moles of } \mathrm{ICl}_3}\) 4. \(\frac{2 \text{ moles of } \mathrm{ICl}_3}{1 \text{ mole of } \mathrm{I}_2}\) 5. \(\frac{3 \text{ moles of } \mathrm{Cl}_2}{2 \text{ moles of } \mathrm{ICl}_3}\) 6. \(\frac{2 \text{ moles of } \mathrm{ICl}_3}{3 \text{ moles of } \mathrm{Cl}_2}\) (c) Atom-based conversion factors: 1. \(\frac{2 \text{ moles of I atoms}}{6 \text{ moles of Cl atoms}}\) 2. \(\frac{6 \text{ moles of Cl atoms}}{2 \text{ moles of I atoms}}\) 3. \(\frac{2 \text{ moles of I atoms}}{6 \text{ moles of ICl molecules}}\) 4. \(\frac{6 \text{ moles of ICl molecules}}{2 \text{ moles of I atoms}}\) 5. \(\frac{6 \text{ moles of Cl atoms}}{4 \text{ moles of ICl molecules}}\) 6. \(\frac{4 \text{ moles of ICl molecules}}{6 \text{ moles of Cl atoms}}\)

Step 1: Write the given (unbalanced) chemical equation

The given chemical equation is: \[\mathrm{I}_{2}+\mathrm{Cl}_{2} \rightarrow \mathrm{ICl}_{3}\]

Step 2: Identify the elements in the equation

We have the following elements in the equation: - Iodine (I) - Chlorine (Cl)

Step 3: Balance the equation for each element

Balance each element by adjusting the coefficients in front of molecules/compounds. Iodine (I): - There are 2 iodine atoms on the left side (LHS) and 2 iodine atoms in 1 mole of \(\mathrm{ICl}_3\) on the right side (RHS). So Iodine is already balanced. Chlorine (Cl): - There are 2 chlorine atoms in 1 mole of \(\mathrm{Cl}_2\) on the left side and 3 chlorine atoms in 1 mole of \(\mathrm{ICl}_3\) on the right side. To balance the Cl atoms, we need to find the least common multiple (LCM) of 2 and 3 which is 6. We can achieve this by putting coefficients of 3 in front of \(\mathrm{Cl}_2\) and 2 in front of \(\mathrm{ICl}_3\): \[ \mathrm{I}_2 + 3\mathrm{Cl}_2 \rightarrow 2\mathrm{ICl}_3\] Now we have a balanced chemical equation. #b. Write all conversion factors containing the two reactants and one product.#

Step 4: Find all mole-based conversion factors

Using the balanced chemical equation, we can find mole-based conversion factors: 1. \(\frac{1 \text{ mole of } \mathrm{I}_2}{3 \text{ moles of } \mathrm{Cl}_2}\) 2. \(\frac{3 \text{ moles of } \mathrm{Cl}_2}{1 \text{ mole of } \mathrm{I}_2}\) 3. \(\frac{1 \text{ mole of } \mathrm{I}_2}{2 \text{ moles of } \mathrm{ICl}_3}\) 4. \(\frac{2 \text{ moles of } \mathrm{ICl}_3}{1 \text{ mole of } \mathrm{I}_2}\) 5. \(\frac{3 \text{ moles of } \mathrm{Cl}_2}{2 \text{ moles of } \mathrm{ICl}_3}\) 6. \(\frac{2 \text{ moles of } \mathrm{ICl}_3}{3 \text{ moles of } \mathrm{Cl}_2}\) #c. Write all conversion factors involving individual atoms and molecules.#

Step 5: Find atom-based conversion factors

We can find atom-based conversion factors using individual atoms from the balanced equation. 1. \(\frac{2 \text{ moles of I atoms}}{6 \text{ moles of Cl atoms}}\) 2. \(\frac{6 \text{ moles of Cl atoms}}{2 \text{ moles of I atoms}}\) 3. \(\frac{2 \text{ moles of I atoms}}{6 \text{ moles of ICl molecules}}\) 4. \(\frac{6 \text{ moles of ICl molecules}}{2 \text{ moles of I atoms}}\) 5. \(\frac{6 \text{ moles of Cl atoms}}{4 \text{ moles of ICl molecules}}\) 6. \(\frac{4 \text{ moles of ICl molecules}}{6 \text{ moles of Cl atoms}}\)

Key Concepts

Mole Conversion Factors

Understanding mole conversion factors is essential for navigating chemical equations. When a chemical equation is balanced, it offers a relationship between moles of reactants and products. These relationships are called mole conversion factors, which help in converting quantities of substances in chemical reactions. For example, consider the balanced equation: \[ \mathrm{I}_2 + 3\mathrm{Cl}_2 \rightarrow 2\mathrm{ICl}_3 \]This equation tells us that:- One mole of \(\mathrm{I}_2\) reacts with three moles of \(\mathrm{Cl}_2\) and forms two moles of \(\mathrm{ICl}_3\).Using these relationships, you can create several mole conversion factors. Here are a few examples:

• \(\frac{1 \text{ mole of } \mathrm{I}_2}{3 \text{ moles of } \mathrm{Cl}_2}\)
• \(\frac{1 \text{ mole of } \mathrm{I}_2}{2 \text{ moles of } \mathrm{ICl}_3}\)

These conversion factors allow chemists to calculate how much of one substance is needed to react with, or produce, another.

Atom-Based Conversion Factors

Atom-based conversion factors take a deeper look at the individual atoms involved in a chemical equation. While mole conversion factors deal with entire molecules, atom-based factors focus on the count of specific atoms participating in the reaction. Using the balanced equation for \(\mathrm{I}_2 + 3\mathrm{Cl}_2 \rightarrow 2\mathrm{ICl}_3\), we can derive atom-based conversion factors. These inform us, for instance, that:

• There are 2 moles of iodine atoms for every 6 moles of chlorine atoms that combine to form \(\mathrm{ICl}_3\).
• \(\frac{6 \text{ moles of Cl atoms}}{2 \text{ moles of I atoms}}\)

These factors are instrumental when calculating the stoichiometry of reactions at the atomic level, ensuring accurate chemical computations.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They depict the substances participating in the reaction (reactants) and those produced (products). Initially, a chemical equation may be unbalanced, such as: \(\mathrm{I}_2 + \mathrm{Cl}_2 \rightarrow \mathrm{ICl}_3\). This unbalanced equation suggests a scenario where the number of atoms on both sides isn’t equal.Balancing equations involves adjusting coefficients to maintain the law of conservation of mass. This means every atom that enters the reaction must leave in some form. A balanced form of the equation looks like: \(\mathrm{I}_2 + 3\mathrm{Cl}_2 \rightarrow 2\mathrm{ICl}_3\), where the count of iodine (I) and chlorine (Cl) atoms is now equivalent on both sides. Balancing ensures accurate and precise reaction stoichiometry.

Reactants and Products

In any chemical equation, understanding the roles of reactants and products is vital. **Reactants** are the substances consumed during the reaction, while **products** are the substances formed.For the chemical reaction represented by \(\mathrm{I}_2 + 3\mathrm{Cl}_2 \rightarrow 2\mathrm{ICl}_3\): - **Reactants**: Iodine molecules \(\mathrm{I}_2\) and chlorine molecules \(\mathrm{Cl}_2\).- **Product**: Iodine chloride \(\mathrm{ICl}_3\).Identifying reactants and products in an equation is the first step in mapping out the reaction process. It allows us to determine what substances are interacting and what compounds they form, which is essential for further analysis and calculations.