Calculating Mass Of Fe2O3 Needed To Produce 100.0 G Fe

Introduction

In the fascinating realm of chemistry, stoichiometry serves as the bedrock for quantitative chemical reactions. It allows us to predict the amounts of reactants and products involved in a chemical transformation, much like a recipe guides a chef in preparing a dish. In this comprehensive article, we will delve into a stoichiometry problem that involves calculating the mass of iron(III) oxide ($Fe_2O_3$) required to produce a specific amount of iron (Fe). This problem not only showcases the practical application of stoichiometric principles but also highlights the importance of mole ratios, molar masses, and balanced chemical equations in chemical calculations. Understanding these concepts is crucial for anyone venturing into the world of chemistry, be it students, researchers, or industry professionals. So, let's embark on this stoichiometric journey and unravel the secrets behind this intriguing chemical calculation.

Problem Statement: The Stoichiometric Puzzle

Our chemical puzzle revolves around the following balanced chemical equation:

$Fe_2O_3(s) + 2Al(s) \rightarrow 2Fe(s) + Al_2O_3(s)$

This equation elegantly describes the reaction between iron(III) oxide ($Fe_2O_3$) and aluminum (Al), resulting in the formation of iron (Fe) and aluminum oxide ($Al_2O_3$). The core of our problem lies in determining the precise mass of $Fe_2O_3$ needed to produce 100.0 g of Fe. This seemingly simple question unlocks a world of stoichiometric calculations and requires a meticulous approach to solve. To successfully navigate this problem, we need to carefully consider the mole ratios between reactants and products, the molar masses of the compounds involved, and the fundamental principles of stoichiometry. This problem serves as an excellent example of how stoichiometry is used to predict the quantities of materials needed or produced in chemical reactions, a cornerstone of chemical engineering, industrial chemistry, and laboratory research. By mastering the techniques required to solve this problem, you'll gain a deeper understanding of how chemical reactions can be precisely controlled and optimized, paving the way for advancements in various scientific and technological fields.

Dissecting the Chemical Equation: Mole Ratios

The first step in conquering this stoichiometry challenge lies in understanding the mole ratios embedded within the balanced chemical equation.

$Fe_2O_3(s) + 2Al(s) \rightarrow 2Fe(s) + Al_2O_3(s)$

This equation reveals the fundamental relationships between the reactants and products at the molecular level. A crucial piece of information we glean from this equation is the mole ratio between $Fe_2O_3$ and Fe. According to the equation, 1 mole of $Fe_2O_3$ reacts to produce 2 moles of Fe. This mole ratio acts as a conversion factor, allowing us to translate between the amount of $Fe_2O_3$ consumed and the amount of Fe produced. Think of it as a recipe: for every 1 "unit" of $Fe_2O_3$, we get 2 "units" of Fe. This mole ratio is not merely a number; it's a fundamental principle that governs the reaction. It dictates the proportions in which reactants combine and products form, ensuring that the law of conservation of mass is upheld. In essence, the mole ratio is the language of stoichiometry, enabling us to quantify the relationships within chemical reactions. Without understanding these ratios, we would be lost in a sea of chemical equations, unable to accurately predict the outcomes of reactions or design efficient chemical processes. This concept is pivotal in various fields, from pharmaceutical synthesis to material science, where precise control over reaction outcomes is paramount.

Calculating Molar Masses: The Bridge Between Mass and Moles

To bridge the gap between grams (the unit we're given for iron) and moles (the language of the chemical equation), we need to calculate the molar masses of $Fe_2O_3$ and Fe. The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). It's a fundamental property that connects the macroscopic world of grams to the microscopic world of atoms and molecules.

Molar Mass of Iron (Fe)

To find the molar mass of Fe, we consult the periodic table. The atomic mass of Fe is approximately 55.845 atomic mass units (amu). Since 1 amu is approximately equal to 1 g/mol, the molar mass of Fe is approximately 55.845 g/mol. This value represents the mass of 6.022 x 10^23 iron atoms (Avogadro's number), which constitutes one mole of iron. The molar mass serves as a conversion factor, allowing us to seamlessly switch between grams of iron and moles of iron. For instance, if we have 111.69 g of iron, we can divide this mass by the molar mass (55.845 g/mol) to find that we have approximately 2 moles of iron.

Molar Mass of Iron(III) Oxide ($Fe_2O_3$)

To calculate the molar mass of $Fe_2O_3$, we need to consider the individual atomic masses of its constituent elements: iron (Fe) and oxygen (O). From the periodic table, we know that the atomic mass of Fe is approximately 55.845 amu, and the atomic mass of O is approximately 16.00 amu. $Fe_2O_3$ contains 2 iron atoms and 3 oxygen atoms. Therefore, the molar mass of $Fe_2O_3$ is calculated as follows:

(2 x 55.845 g/mol) + (3 x 16.00 g/mol) = 159.69 g/mol

This molar mass tells us that one mole of $Fe_2O_3$ weighs approximately 159.69 grams. This value is essential for converting between grams of $Fe_2O_3$ and moles of $Fe_2O_3$, a crucial step in solving our stoichiometry problem. Just like the molar mass of iron, the molar mass of $Fe_2O_3$ acts as a bridge, connecting the tangible world of grams to the abstract world of moles. By understanding and utilizing molar masses, we can perform accurate chemical calculations, predict reaction outcomes, and design chemical processes with confidence.

The Stoichiometric Calculation: A Step-by-Step Guide

Now that we have the necessary tools – the mole ratio and the molar masses – we can embark on the stoichiometric calculation to determine the mass of $Fe_2O_3$ required to produce 100.0 g of Fe. We will approach this problem in a step-by-step manner, ensuring clarity and accuracy.

Step 1: Convert Grams of Fe to Moles of Fe

We begin by converting the given mass of Fe (100.0 g) to moles using the molar mass of Fe (55.845 g/mol):

Moles of Fe = (100.0 g Fe) / (55.845 g/mol Fe) ≈ 1.791 moles Fe

This conversion allows us to express the amount of iron in terms of moles, which is the language of the balanced chemical equation. We now know that we need to produce approximately 1.791 moles of iron.

Step 2: Use the Mole Ratio to Find Moles of $Fe_2O_3$

Next, we employ the mole ratio from the balanced chemical equation to determine the number of moles of $Fe_2O_3$ required. Recall that the equation tells us that 1 mole of $Fe_2O_3$ produces 2 moles of Fe. We can use this ratio as a conversion factor:

Moles of $Fe_2O_3$ = (1.791 moles Fe) x (1 mole $Fe_2O_3$ / 2 moles Fe) ≈ 0.8955 moles $Fe_2O_3$

This calculation reveals that we need approximately 0.8955 moles of $Fe_2O_3$ to produce 1.791 moles of Fe. The mole ratio has acted as a bridge, connecting the amount of product (Fe) to the amount of reactant ($Fe_2O_3$).

Step 3: Convert Moles of $Fe_2O_3$ to Grams of $Fe_2O_3$

Finally, we convert the moles of $Fe_2O_3$ to grams using the molar mass of $Fe_2O_3$ (159.69 g/mol):

Mass of $Fe_2O_3$ = (0.8955 moles $Fe_2O_3$) x (159.69 g/mol $Fe_2O_3$) ≈ 143.0 g $Fe_2O_3$

This final calculation provides the answer to our problem: we need approximately 143.0 grams of $Fe_2O_3$ to produce 100.0 grams of Fe. This step completes the stoichiometric calculation, transforming the abstract concept of moles back into a tangible mass that can be measured in the laboratory.

The Answer: 143.0 g of $Fe_2O_3$

Therefore, the mass of $Fe_2O_3$ that must be used to produce 100.0 g of Fe is approximately 143.0 g. This corresponds to answer choice A in the given options.

Key Takeaways: Mastering Stoichiometry

This problem exemplifies the core principles of stoichiometry and highlights the importance of:

• Balanced Chemical Equations: The foundation of all stoichiometric calculations, providing the crucial mole ratios.
• Mole Ratios: The bridge between reactants and products, allowing us to predict the quantities involved in a reaction.
• Molar Masses: The link between grams and moles, enabling us to convert between mass and the number of particles.
• Step-by-Step Problem Solving: Breaking down complex problems into manageable steps for clarity and accuracy.

By mastering these concepts, you can confidently tackle a wide range of stoichiometry problems and gain a deeper understanding of the quantitative nature of chemical reactions. Stoichiometry is not just a set of calculations; it's a powerful tool that allows us to predict, control, and optimize chemical processes in various fields, from medicine to manufacturing. Whether you're a student learning the basics or a professional designing chemical plants, a solid grasp of stoichiometry is indispensable.

Further Practice: Sharpening Your Stoichiometry Skills

To solidify your understanding of stoichiometry, practice is key. Try solving similar problems with different reactants, products, and target masses. Explore variations such as limiting reactant problems or percent yield calculations. The more you practice, the more comfortable and confident you'll become in applying these principles. Consider exploring online resources, textbooks, and practice worksheets to expand your knowledge and challenge your skills. Stoichiometry is a skill that builds upon itself, so consistent practice will lead to mastery. Don't be afraid to make mistakes; they are valuable learning opportunities. Analyze your errors, identify areas for improvement, and keep practicing. With dedication and perseverance, you can become a stoichiometry expert!

Conclusion: Stoichiometry – The Language of Chemistry

In conclusion, the problem of determining the mass of $Fe_2O_3$ needed to produce 100.0 g of Fe beautifully illustrates the power and elegance of stoichiometry. By carefully applying the principles of mole ratios, molar masses, and balanced chemical equations, we successfully navigated the calculation and arrived at the correct answer. Stoichiometry is more than just a set of formulas and equations; it's the language of chemistry, allowing us to quantify and understand the relationships between substances in chemical reactions. Mastering stoichiometry is essential for anyone pursuing a career in chemistry or related fields, and it provides a fundamental understanding of how the world around us works at a molecular level. So, embrace the challenges, practice the techniques, and unlock the secrets of stoichiometry – the key to understanding the quantitative world of chemistry.