Calculating Moles Of Iron(III) Sulfide Produced From Iron(III) Bromide

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Determining the stoichiometry of chemical reactions is a fundamental aspect of chemistry. In this comprehensive guide, we will walk through the process of calculating the moles of iron(III) sulfide ($Fe_2S_3$) produced from the complete reaction of 449 g of iron(III) bromide ($FeBr_3$). We will break down the steps, explain the underlying concepts, and provide a clear, step-by-step solution. This article is designed for students, educators, and chemistry enthusiasts looking to enhance their understanding of stoichiometry.

Understanding Stoichiometry

Stoichiometry, derived from the Greek words stoicheion (element) and metron (measure), is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It is crucial for predicting the amounts of reactants needed and products formed in a chemical reaction. Stoichiometric calculations are based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This means the number of atoms of each element must be the same on both sides of the chemical equation.

Key Concepts in Stoichiometry

1. Chemical Equations: A balanced chemical equation is the foundation of stoichiometry. It provides the molar ratios between reactants and products. For instance, in our reaction:

   $$2 FeBr_3 + 3 Na_2S$$

ightarrow Fe_2S_3 + 6 NaBr$

This equation tells us that 2 moles of $FeBr_3$ react with 3 moles of $Na_2S$ to produce 1 mole of $Fe_2S_3$ and 6 moles of $NaBr$.

2. Moles: The mole is the SI unit for the amount of substance. One mole contains Avogadro's number ($6.022 imes 10^{23}$) of entities (atoms, molecules, ions, etc.).
3. Molar Mass: The molar mass of a substance is the mass of one mole of that substance, usually expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in the chemical formula.
4. Stoichiometric Ratio: The stoichiometric ratio is the ratio of the coefficients of reactants and products in a balanced chemical equation. It allows us to convert between moles of different substances in a reaction.

Problem Statement

Our primary objective is to determine how many moles of iron(III) sulfide ($Fe_2S_3$) will be produced from the complete reaction of 449 g of iron(III) bromide ($FeBr_3$). The balanced chemical equation for the reaction is:

$$2 FeBr_3 + 3 Na_2S ightarrow Fe_2S_3 + 6 NaBr$$

To solve this problem, we will follow these steps:

1. Calculate the molar mass of $FeBr_3$.
2. Convert the mass of $FeBr_3$ to moles.
3. Use the stoichiometric ratio from the balanced equation to find the moles of $Fe_2S_3$ produced.

Step-by-Step Solution

Step 1: Calculate the Molar Mass of Iron(III) Bromide ($FeBr_3$)

The molar mass of a compound is the sum of the atomic masses of each element in the compound, multiplied by their respective subscripts. For $FeBr_3$, we have:

• Iron (Fe): 1 atom × 55.845 g/mol = 55.845 g/mol
• Bromine (Br): 3 atoms × 79.904 g/mol = 239.712 g/mol

Adding these values together:

Molar mass of $FeBr_3$ = 55.845 g/mol + 239.712 g/mol = 295.557 g/mol

Therefore, the molar mass of $FeBr_3$ is approximately 295.557 g/mol.

Step 2: Convert the Mass of Iron(III) Bromide ($FeBr_3$) to Moles

To convert the mass of $FeBr_3$ to moles, we use the formula:

ext{Moles} = rac{ ext{Mass}}{ ext{Molar Mass}}

Given mass of $FeBr_3$ = 449 g Molar mass of $FeBr_3$ = 295.557 g/mol

ext{Moles of } FeBr_3 = rac{449 ext{ g}}{295.557 ext{ g/mol}} \[0.3cm] ext{Moles of } FeBr_3 ext{ ≈ } 1.52 ext{ mol}

Thus, we have approximately 1.52 moles of $FeBr_3$.

Step 3: Use the Stoichiometric Ratio to Find Moles of Iron(III) Sulfide ($Fe_2S_3$) Produced

From the balanced chemical equation:

$$2 FeBr_3 + 3 Na_2S ightarrow Fe_2S_3 + 6 NaBr$$

We see that 2 moles of $FeBr_3$ produce 1 mole of $Fe_2S_3$. The stoichiometric ratio between $FeBr_3$ and $Fe_2S_3$ is 2:1. We can use this ratio to convert moles of $FeBr_3$ to moles of $Fe_2S_3$:

ext{Moles of } Fe_2S_3 = ext{Moles of } FeBr_3 imes rac{ ext{Moles of } Fe_2S_3}{ ext{Moles of } FeBr_3}

ext{Moles of } Fe_2S_3 = 1.52 ext{ mol } FeBr_3 imes rac{1 ext{ mol } Fe_2S_3}{2 ext{ mol } FeBr_3} \[0.3cm] ext{Moles of } Fe_2S_3 = 1.52 ext{ mol } FeBr_3 imes 0.5 \[0.3cm] ext{Moles of } Fe_2S_3 ext{ ≈ } 0.76 ext{ mol}

Therefore, approximately 0.76 moles of $Fe_2S_3$ will be produced from the complete reaction of 449 g of $FeBr_3$.

Conclusion

In summary, we have calculated that the complete reaction of 449 g of iron(III) bromide ($FeBr_3$) will produce approximately 0.76 moles of iron(III) sulfide ($Fe_2S_3$). This calculation involved understanding and applying stoichiometric principles, including molar mass calculations and the use of stoichiometric ratios from the balanced chemical equation.

Stoichiometry is a critical tool in chemistry, allowing us to predict the quantities of substances involved in chemical reactions accurately. By following a systematic approach, we can solve complex stoichiometric problems with confidence. This problem illustrates the importance of converting masses to moles and utilizing the mole ratios from balanced chemical equations to determine the amount of products formed.

The ability to perform these calculations is essential for various applications, from laboratory experiments to industrial chemical processes. Mastering stoichiometry provides a solid foundation for further studies in chemistry and related fields.

Further Practice

To reinforce your understanding of stoichiometry, consider working through additional practice problems. Here are a few suggestions:

1. Calculate the mass of $Na_2S$ required to react completely with 449 g of $FeBr_3$ in the given reaction.
2. Determine the mass of $NaBr$ produced in the reaction.
3. Explore other stoichiometric problems involving different chemical reactions and compounds.

By practicing these types of problems, you will enhance your skills in stoichiometry and deepen your understanding of chemical reactions.

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Frequently Asked Questions (FAQs)

1. What is stoichiometry? Stoichiometry is the branch of chemistry that involves the quantitative relationships between reactants and products in chemical reactions. It is based on the law of conservation of mass and allows us to predict the amounts of substances involved in a reaction.

2. Why is stoichiometry important? Stoichiometry is crucial for accurately predicting the quantities of reactants needed and products formed in chemical reactions. It is essential in various applications, including laboratory experiments, industrial chemical processes, and environmental chemistry.

3. What is a mole? A mole is the SI unit for the amount of substance. One mole contains Avogadro's number ($6.022 imes 10^{23}$) of entities (atoms, molecules, ions, etc.).

4. How do you calculate molar mass? The molar mass of a substance is calculated by summing the atomic masses of all the atoms in the chemical formula. The atomic masses are typically found on the periodic table.

5. What is a balanced chemical equation? A balanced chemical equation is a chemical equation in which the number of atoms of each element is the same on both sides of the equation. Balancing chemical equations is essential for stoichiometric calculations.

6. How do you use stoichiometric ratios? Stoichiometric ratios are derived from the coefficients in a balanced chemical equation. They allow us to convert between moles of different substances in a reaction. For example, if the ratio between two substances is 2:1, then 2 moles of the first substance react with 1 mole of the second substance.

7. What is the law of conservation of mass? The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. This means the total mass of the reactants must equal the total mass of the products.

8. How do you convert grams to moles? To convert grams to moles, use the formula: $ ext{Moles} = rac{ ext{Mass}}{ ext{Molar Mass}}$. Divide the mass of the substance by its molar mass.

9. What are the steps to solving stoichiometry problems? The general steps for solving stoichiometry problems are:

   1. Write and balance the chemical equation.
   2. Convert given masses to moles.
   3. Use the stoichiometric ratio to find moles of the desired substance.
   4. Convert moles back to grams if needed.

10. Can stoichiometry be applied to real-world scenarios? Yes, stoichiometry is widely applied in various real-world scenarios, such as:

    • Industrial Chemistry: Optimizing chemical processes to maximize product yield and minimize waste.
    • Pharmaceutical Industry: Calculating the amounts of reactants needed for drug synthesis.
    • Environmental Science: Determining the amount of pollutants produced in chemical reactions.
    • Cooking: Adjusting ingredient quantities based on the number of servings.

By understanding these FAQs, you can reinforce your knowledge of stoichiometry and its applications, making you better equipped to tackle more complex chemical problems.

Conclusion Revisited

In conclusion, the journey through the world of stoichiometry, particularly in calculating the moles of iron(III) sulfide ($Fe_2S_3$) produced from the complete reaction of 449 g of iron(III) bromide ($FeBr_3$), reveals the elegance and precision of chemical calculations. This exploration not only reinforces the foundational principles of chemistry but also highlights the practical applications of these concepts in real-world scenarios. Stoichiometry, at its core, is the art and science of quantifying chemical reactions, enabling us to predict outcomes and optimize processes.

The Importance of Precision in Stoichiometric Calculations

Throughout our step-by-step solution, we emphasized the significance of precision in calculations. Each step, from determining molar masses to applying stoichiometric ratios, requires careful attention to detail. The molar mass calculation, for instance, is the bedrock upon which the entire calculation rests. A slight error in molar mass can propagate through subsequent steps, leading to a significant deviation in the final result. Similarly, the stoichiometric ratio derived from the balanced chemical equation serves as the bridge between reactants and products. An accurate ratio ensures the correct proportion of substances is considered, reflecting the true chemical transformation occurring.

The Role of the Balanced Chemical Equation

The balanced chemical equation is more than just a symbolic representation of a chemical reaction; it is the cornerstone of stoichiometric calculations. It provides the essential mole ratios that dictate how reactants transform into products. In our specific problem, the equation $2 FeBr_3 + 3 Na_2S ightarrow Fe_2S_3 + 6 NaBr$ tells us that two moles of iron(III) bromide react to produce one mole of iron(III) sulfide. This 2:1 ratio is the key to converting moles of the reactant to moles of the product. Without a balanced equation, stoichiometric calculations would be akin to navigating without a map, leading to inaccurate and unreliable results.

Real-World Applications and Implications

The applications of stoichiometry extend far beyond the classroom and the laboratory. In industrial chemistry, stoichiometry is vital for optimizing reaction yields and minimizing waste. Chemical engineers rely on stoichiometric calculations to determine the precise amounts of reactants needed for a desired product output. In the pharmaceutical industry, the synthesis of drugs requires meticulous stoichiometric planning to ensure the correct proportions of reactants are used, leading to the desired therapeutic compound. Environmental scientists employ stoichiometry to assess pollution levels, track chemical transformations in the environment, and develop strategies for remediation. Even in culinary arts, stoichiometry finds its place, as chefs adjust ingredient quantities to maintain the integrity of their recipes.

Building a Strong Foundation in Chemistry

Mastering stoichiometry is not merely about solving numerical problems; it is about building a strong foundation in chemistry. The concepts learned through stoichiometry—moles, molar mass, balanced equations, and mole ratios—are fundamental to understanding chemical reactions and the behavior of matter. This foundational knowledge paves the way for exploring more advanced topics, such as chemical kinetics, thermodynamics, and equilibrium. By understanding how substances react quantitatively, we gain deeper insights into the molecular world and the forces that govern it.

Continuing the Journey of Learning

The journey of learning stoichiometry is ongoing. As we delve deeper into chemistry, we encounter more complex reactions and intricate calculations. However, the basic principles of stoichiometry remain constant, providing a reliable framework for solving a wide range of problems. By engaging with practice problems, exploring real-world applications, and seeking clarification on challenging concepts, we can strengthen our grasp of stoichiometry and unlock the fascinating world of chemical transformations.

In closing, the calculation of moles of iron(III) sulfide from the reaction of iron(III) bromide is a testament to the power and precision of stoichiometry. It underscores the importance of meticulous calculations, balanced chemical equations, and a deep understanding of chemical principles. As we continue our exploration of chemistry, stoichiometry will undoubtedly serve as a valuable tool, enabling us to quantify the world around us and make meaningful contributions to science and society.