Stoichiometry Calculation How Much Strontium Sulfide Is Needed To Produce 209 G Of Sulfur

Introduction: The Stoichiometry of Strontium Sulfide and Sulfur

In the realm of chemistry, understanding the relationships between reactants and products in a chemical reaction is paramount. This intricate dance of atoms and molecules, governed by the laws of stoichiometry, allows us to predict and quantify the outcomes of chemical processes. In this article, we delve into a specific scenario: a scientist's quest to determine the precise mass of strontium sulfide (SrS) required to produce 209 grams of sulfur (S₈). This exploration will not only illuminate the practical applications of stoichiometry but also highlight the fundamental principles that underpin chemical transformations.

The ability to precisely calculate the mass of reactants needed to obtain a desired amount of product is crucial in various fields, from industrial manufacturing to laboratory research. Whether it's synthesizing new materials, optimizing chemical processes, or simply performing experiments, a solid grasp of stoichiometry is indispensable. Our focus on the strontium sulfide-sulfur reaction serves as a compelling case study, showcasing the step-by-step process of stoichiometric calculations and the importance of balanced chemical equations. We will embark on a journey through the world of molar masses, mole ratios, and the art of converting between grams and moles, ultimately unraveling the secrets behind this chemical transformation. This understanding is not just academic; it's a practical skill that empowers chemists and scientists to manipulate matter at the molecular level.

Balancing the Chemical Equation: Laying the Foundation

The first step in any stoichiometric calculation is to ensure that the chemical equation representing the reaction is properly balanced. A balanced chemical equation is a symbolic representation of a chemical reaction that adheres to the law of conservation of mass. This law dictates that matter cannot be created or destroyed in a chemical reaction, meaning the number of atoms of each element must be the same on both the reactant and product sides of the equation. Balancing equations is not merely a matter of aesthetics; it's a fundamental requirement for accurate stoichiometric calculations.

The unbalanced equation for the reaction between strontium sulfide (SrS) and oxygen (O₂) to produce strontium oxide (SrO) and sulfur (S₈) is:

SrS + O₂ → SrO + S₈

At first glance, it's evident that the equation is unbalanced. The sulfur atoms are the most glaring discrepancy, with eight sulfur atoms on the product side (S₈) and only one on the reactant side (SrS). To rectify this, we begin by placing a coefficient of 8 in front of SrS:

8 SrS + O₂ → SrO + S₈

This adjustment balances the sulfur atoms, but now the strontium atoms are unbalanced. There are eight strontium atoms on the reactant side (8 SrS) and only one on the product side (SrO). To balance the strontium, we place a coefficient of 8 in front of SrO:

8 SrS + O₂ → 8 SrO + S₈

Now, the strontium and sulfur atoms are balanced, but the oxygen atoms remain unbalanced. There are two oxygen atoms on the reactant side (O₂) and eight oxygen atoms on the product side (8 SrO). To balance the oxygen, we place a coefficient of 8 in front of O₂:

8 SrS + 8 O₂ → 8 SrO + S₈

With this final adjustment, the equation is now balanced. The number of atoms of each element is the same on both sides of the equation, satisfying the law of conservation of mass. This balanced equation serves as the foundation for all subsequent stoichiometric calculations, providing the crucial mole ratios that link reactants and products.

Calculating Molar Masses: Bridging Grams and Moles

Stoichiometry operates in the realm of moles, a unit that represents a specific number of particles (6.022 x 10²³ particles, Avogadro's number). However, in the laboratory, we typically measure masses in grams. Therefore, to navigate between the macroscopic world of grams and the microscopic world of moles, we need to employ molar masses. The molar mass of a substance is the mass of one mole of that substance, expressed in grams per mole (g/mol). It serves as a crucial conversion factor, allowing us to translate between mass and the number of moles.

To calculate the molar mass of a compound, we sum the atomic masses of all the atoms present in its chemical formula. These atomic masses can be obtained from the periodic table. For strontium sulfide (SrS), the molar mass calculation proceeds as follows:

• Molar mass of Sr = 87.62 g/mol
• Molar mass of S = 32.07 g/mol
• Molar mass of SrS = Molar mass of Sr + Molar mass of S
• Molar mass of SrS = 87.62 g/mol + 32.07 g/mol
• Molar mass of SrS = 119.69 g/mol

Thus, the molar mass of strontium sulfide (SrS) is 119.69 g/mol. This value signifies that one mole of SrS weighs 119.69 grams. Similarly, we calculate the molar mass of sulfur (S₈):

• Molar mass of S = 32.07 g/mol
• Molar mass of S₈ = 8 * Molar mass of S
• Molar mass of S₈ = 8 * 32.07 g/mol
• Molar mass of S₈ = 256.56 g/mol

The molar mass of sulfur (S₈) is 256.56 g/mol, indicating that one mole of S₈ weighs 256.56 grams. These molar masses are essential tools in our stoichiometric calculations, acting as bridges between the mass of a substance and the number of moles it represents. With these values in hand, we can confidently convert between grams and moles, a crucial step in determining the mass of SrS required to produce 209 grams of S₈.

Stoichiometric Calculations: From Grams of Sulfur to Grams of Strontium Sulfide

With the balanced equation and molar masses established, we can now embark on the core of the problem: determining the mass of strontium sulfide (SrS) needed to produce 209 grams of sulfur (S₈). This involves a series of conversions, guided by the principles of stoichiometry and the information encoded within the balanced chemical equation. The balanced equation, 8 SrS + 8 O₂ → 8 SrO + S₈, reveals the crucial mole ratio between SrS and S₈. For every one mole of S₈ produced, eight moles of SrS are consumed. This mole ratio is the key to unlocking the solution.

The calculation proceeds in a step-by-step manner, ensuring that units cancel out appropriately, leading us to the desired result. First, we convert the given mass of sulfur (209 grams) into moles using its molar mass (256.56 g/mol):

Moles of S₈ = (Mass of S₈) / (Molar mass of S₈) Moles of S₈ = (209 g) / (256.56 g/mol) Moles of S₈ ≈ 0.815 mol

Next, we utilize the mole ratio from the balanced equation to determine the moles of SrS required. Since 8 moles of SrS are needed to produce 1 mole of S₈, we multiply the moles of S₈ by this ratio:

Moles of SrS = (Moles of S₈) * (8 mol SrS / 1 mol S₈) Moles of SrS = (0.815 mol) * (8 mol SrS / 1 mol S₈) Moles of SrS ≈ 6.52 mol

Finally, we convert the moles of SrS back into grams using its molar mass (119.69 g/mol):

Mass of SrS = (Moles of SrS) * (Molar mass of SrS) Mass of SrS = (6.52 mol) * (119.69 g/mol) Mass of SrS ≈ 780.3 g

Therefore, approximately 780.3 grams of strontium sulfide (SrS) are needed to produce 209 grams of sulfur (S₈). This calculation demonstrates the power of stoichiometry in predicting the quantitative relationships between reactants and products in a chemical reaction. By carefully applying the principles of balancing equations, molar masses, and mole ratios, we can confidently navigate the world of chemical transformations and achieve our desired outcomes.

Conclusion: The Significance of Stoichiometry in Chemical Synthesis

In conclusion, the determination of the mass of strontium sulfide (SrS) required to produce 209 grams of sulfur (S₈) exemplifies the fundamental role of stoichiometry in chemical synthesis and quantitative analysis. This exercise underscores the importance of balanced chemical equations, molar masses, and mole ratios in accurately predicting the outcomes of chemical reactions. Stoichiometry is not merely a theoretical concept; it is a practical tool that empowers chemists and scientists to manipulate matter at the molecular level, enabling them to design and optimize chemical processes.

The ability to calculate reactant quantities needed for a desired product yield is crucial in various applications, ranging from industrial manufacturing to pharmaceutical development. Stoichiometry allows us to minimize waste, maximize efficiency, and ensure the cost-effective production of chemicals and materials. Furthermore, a solid understanding of stoichiometry is essential for interpreting experimental data, analyzing reaction mechanisms, and developing new chemical reactions. It is a cornerstone of chemical knowledge, providing a framework for understanding and predicting the behavior of matter.

The specific example of strontium sulfide and sulfur production highlights the step-by-step approach to stoichiometric calculations. From balancing the chemical equation to converting between grams and moles, each step is critical in arriving at the correct answer. The use of molar masses as conversion factors and the application of mole ratios from the balanced equation are key techniques that can be applied to a wide range of stoichiometric problems. By mastering these principles, students and professionals alike can confidently tackle complex chemical challenges and contribute to advancements in various scientific and technological fields. The journey from a seemingly simple question – how much strontium sulfide is needed? – to a precise answer underscores the power and elegance of stoichiometry in the world of chemistry.