Problem 52 - Chemistry The Molecular Science

Question

Determine the amount of reaction (in moles) that takes place for each process $$ \frac{1}{2} \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) \longrightarrow \mathrm{FeO}(\mathrm{s})+\frac{1}{4} \mathrm{O}_{2}(\mathrm{~g}) $$ (a) \(2 \mathrm{~mol} \mathrm{O}_{2}\) forms (b) \(0.824 \mathrm{~mol} \mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts (c) \(1.34 \mathrm{~g} \mathrm{FeO}\) forms

Step-by-Step Solution

Answer

(a) 4 mol, (b) 0.824 mol, (c) 0.009325 mol.

Step 1: Determine Moles from Stoichiometry (Part a)

For part (a), we start with the formation of \(2 \text{ mol } \mathrm{O}_2\). From the balanced equation, for every \(\frac{1}{4} \text{ mol } \mathrm{O}_2\) formed, \(\frac{1}{2} \text{ mol } \mathrm{Fe}_2\mathrm{O}_3\) reacts. Therefore, for \(2 \text{ mol } \mathrm{O}_2\):\[\frac{2}{\frac{1}{4}} \times \frac{1}{2} = 4 \text{ mol } \mathrm{Fe}_2\mathrm{O}_3\] Thus, 4 moles of \(\mathrm{Fe}_2\mathrm{O}_3\) react.

Step 2: Direct Mole Relation (Part b)

For part (b), we have \(0.824 \text{ mol } \mathrm{Fe}_2\mathrm{O}_3\) reacting directly as given. Therefore, the amount of reaction is simply the same, since we have \(\frac{1}{2} \text{ mol } \mathrm{Fe}_2\mathrm{O}_3\) reacting in this stoichiometry.Hence, 0.824 moles of reaction takes place.

Step 3: Convert Grams to Moles (Part c)

For part (c), first we need to convert grams of \(\mathrm{FeO}\) to moles. The molar mass of \(\mathrm{FeO}\) is approximately 71.85 g/mol. From the equation:\[\text{Moles of } \mathrm{FeO} = \frac{1.34 \text{ g}}{71.85 \text{ g/mol}} \approx 0.01865 \text{ mol}\]According to the equation, for every \(1 \text{ mol } \mathrm{FeO}\) formed, \(\frac{1}{2} \text{ mol } \mathrm{Fe}_2\mathrm{O}_3\) reacts: \[0.01865 \text{ mol } \times \frac{1}{2} \approx 0.009325 \text{ mol}\]Thus, the reaction extent is approximately 0.009325 moles.

Key Concepts

Chemical Reactions

Chemical reactions describe how substances change during a chemical process. They are expressed using chemical equations that show the reactants (starting materials) and products (end results). In the given exercise, the equation is balanced, meaning that the number of atoms of each element is the same on both sides.
Balanced equations follow the law of conservation of mass, ensuring mass is neither created nor destroyed. This provides a clear roadmap for understanding how much of each substance is involved. In our example, for every \(\frac{1}{2}\) mole of \(\mathrm{Fe}_2\mathrm{O}_3\) reacting, \(1\) mole of \(\mathrm{FeO}\) and \(\frac{1}{4}\) mole of \(\mathrm{O}_2\) is formed.

• Reactants are the substances you start with. Here, it's \(\mathrm{Fe}_2\mathrm{O}_3\).
• Products are what you end up with, like \(\mathrm{FeO}\) and \(\mathrm{O}_2\) in this case.

Understanding the relationships between reactants and products is crucial for predicting how much of each substance you'll need or produce in an experiment.

Molar Mass

Molar mass is a vital concept for calculating amounts of substances in chemical reactions. It represents the mass of one mole of a substance, typically expressed in grams per mole (g/mol).
To find the molar mass, sum the atomic masses of all atoms in a molecule, which you can find on the periodic table. For example:

• The molar mass of \(\mathrm{FeO}\) is calculated by adding the atomic masses of iron (Fe) and oxygen (O), giving approximately 71.85 g/mol.

Molar mass acts as a bridge, translating between grams and moles. When you're given mass in grams, you can use molar mass to convert it to moles, a step crucial in stoichiometric calculations.
In the exercise, the molar mass of \(\mathrm{FeO}\) helps convert 1.34 g of \(\mathrm{FeO}\) into moles, making it possible to determine the amount of reaction occurring.

Conversion of Grams to Moles

Converting grams to moles is essential for solving stoichiometry problems. This conversion allows you to translate a measurable quantity (mass) into a countable quantity (moles), which is used in chemical equations.
Here’s how you do it:

• Identify the mass of the substance in grams.
• Use the molar mass (from the periodic table) to convert grams to moles using the formula: \[\text{Moles} = \frac{\text{grams}}{\text{molar mass in g/mol}}\]

In the step-by-step solution, 1.34 g of \(\mathrm{FeO}\) is converted to moles using its molar mass (71.85 g/mol), resulting in approximately 0.01865 moles.
This conversion reveals how much of a reactant is involved, allowing you to use stoichiometry to find out other amounts in the reaction. It simplifies the complex process of transformation in reactions.