Iron Sulfide Formation: Stoichiometry And Limiting Reactants
Hey guys! Let's dive into a fun chemistry problem involving the formation of iron(II) sulfide (FeS). This is a classic stoichiometry question where we need to figure out how much product we can make from given amounts of reactants, and it involves identifying the limiting reactant. So, grab your calculators, and let's get started!
Understanding the Basics of Iron Sulfide (FeS) Formation
To begin, iron sulfide (FeS) forms when iron (Fe) reacts with sulfur (S). The balanced chemical equation for this reaction is:
Fe + S → FeS
From the problem statement, we know that 32 g of sulfur react with 55.8 g of iron. This is crucial information because it tells us the stoichiometric ratio of the reactants. In other words, it tells us the exact amount of each reactant needed to fully react and form the product without any leftovers. Understanding this ratio is key to solving the problem.
Converting Grams to Moles
To truly understand the reaction, we need to convert grams to moles. The molar mass of iron (Fe) is approximately 55.8 g/mol, and the molar mass of sulfur (S) is approximately 32 g/mol. This conversion is important because chemical reactions occur on a mole basis, not a mass basis. So, let's calculate the moles:
- Moles of Fe = 55.8 g / 55.8 g/mol = 1 mol
- Moles of S = 32 g / 32 g/mol = 1 mol
This confirms that iron and sulfur react in a 1:1 molar ratio to form FeS. Now that we have a solid understanding of the basic stoichiometry, we can move on to the actual problem.
Problem Part A: Determining the Amount of FeS Formed
In part (a), we're asked to determine how many grams of FeS will be formed when 30 g of sulfur react with 40 g of iron. This is a limiting reactant problem, meaning one of the reactants will be completely consumed before the other, thus limiting the amount of product that can be formed.
Identifying the Limiting Reactant
To identify the limiting reactant, we need to calculate how many moles of each reactant we have:
- Moles of S = 30 g / 32 g/mol ≈ 0.9375 mol
- Moles of Fe = 40 g / 55.8 g/mol ≈ 0.7168 mol
Since iron and sulfur react in a 1:1 molar ratio, we can see that we have less iron (0.7168 mol) than sulfur (0.9375 mol). This means iron is the limiting reactant, and the amount of FeS formed will be determined by the amount of iron available.
Calculating the Amount of FeS Formed
Since 0.7168 mol of iron will react completely, we will form 0.7168 mol of FeS. Now we need to convert this back to grams. The molar mass of FeS is the sum of the molar masses of iron and sulfur:
- Molar mass of FeS = 55.8 g/mol (Fe) + 32 g/mol (S) = 87.8 g/mol
Now, calculate the mass of FeS formed:
- Mass of FeS = 0.7168 mol * 87.8 g/mol ≈ 62.93 g
So, when 30 g of sulfur react with 40 g of iron, approximately 62.93 grams of FeS will be formed. Remember, the limiting reactant (iron in this case) dictates the maximum amount of product that can be produced.
Problem Part B: Determining the Amount of Fe and S Remaining
In part (b), we need to determine how many grams of Fe and S will be left over after the reaction. We already know that iron is the limiting reactant, so all 40 g of iron will be consumed. Now, let's figure out how much sulfur will be left over.
Calculating Sulfur Used
Since 0.7168 mol of iron reacted, it also means that 0.7168 mol of sulfur reacted. We can calculate the mass of sulfur that reacted:
- Mass of S reacted = 0.7168 mol * 32 g/mol ≈ 22.94 g
Calculating Sulfur Remaining
We started with 30 g of sulfur, and 22.94 g of sulfur reacted. To find out how much sulfur is left over, we subtract the amount reacted from the initial amount:
- Mass of S remaining = 30 g - 22.94 g ≈ 7.06 g
So, after the reaction, approximately 7.06 grams of sulfur will be left over. Since iron was the limiting reactant, there will be no iron left over.
Summary of Results
To summarize:
- Grams of FeS formed: ≈ 62.93 g
- Grams of Fe remaining: 0 g
- Grams of S remaining: ≈ 7.06 g
Deep Dive: Stoichiometry and Limiting Reactants
Stoichiometry is the study of the quantitative relationships or ratios between two or more substances undergoing a physical change or chemical reaction. In simpler terms, it's like a recipe for a chemical reaction. It tells you how much of each ingredient (reactant) you need to make a certain amount of the final dish (product).
Limiting Reactant: In any chemical reaction, the limiting reactant is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reactant since the reaction cannot continue once it's all used up. Identifying the limiting reactant is crucial in determining the maximum yield of the product.
Importance of Understanding Molar Ratios
Understanding molar ratios is vital in stoichiometry. These ratios are derived from the balanced chemical equation and indicate the proportion in which reactants combine and products form. For instance, in our FeS example (Fe + S → FeS), the molar ratio of Fe to S to FeS is 1:1:1. This means one mole of iron reacts with one mole of sulfur to produce one mole of iron sulfide. Ignoring these ratios can lead to incorrect calculations and an inaccurate understanding of the reaction.
Practical Applications of Stoichiometry
Stoichiometry isn't just a theoretical concept; it has numerous practical applications in various fields:
- Pharmaceutical Industry: Stoichiometry is crucial in drug development and manufacturing. It ensures that the right amounts of reactants are used to synthesize drug compounds, guaranteeing the drug's efficacy and safety.
- Chemical Engineering: In chemical plants, stoichiometry is used to optimize chemical reactions for maximum yield and efficiency. Engineers use stoichiometric calculations to design reactors and control processes.
- Environmental Science: Stoichiometry helps in understanding and mitigating environmental pollution. For example, it is used to calculate the amount of chemicals needed to neutralize pollutants in water or air.
- Food Industry: In food production, stoichiometry is used to ensure the correct proportions of ingredients in food products, maintaining consistent quality and nutritional value.
Common Mistakes to Avoid
When working with stoichiometry, it's easy to make mistakes if you're not careful. Here are some common pitfalls to avoid:
- Not Balancing the Chemical Equation: Always make sure the chemical equation is balanced before performing any calculations. An unbalanced equation will lead to incorrect molar ratios.
- Forgetting to Convert to Moles: Chemical reactions occur on a mole basis, so always convert grams to moles before using stoichiometric ratios.
- Incorrectly Identifying the Limiting Reactant: Double-check your calculations to ensure you've correctly identified the limiting reactant, as it determines the amount of product formed.
- Rounding Errors: Be mindful of rounding errors, especially in multi-step calculations. Keep as many significant figures as possible until the final answer.
Advanced Stoichiometry Problems
For those looking to challenge themselves further, here are some advanced stoichiometry problem types:
- Reactions with Multiple Reactants and Products: These involve more complex balanced equations and require careful tracking of molar ratios.
- Reactions with Impure Reactants: When reactants are not 100% pure, you need to account for the percentage of the actual reacting substance.
- Reactions with Percentage Yield: In real-world scenarios, the actual yield of a reaction may be less than the theoretical yield. Percentage yield calculations account for this discrepancy.
- Reactions in Solution: When reactions occur in solution, you need to use molarity and volume to determine the number of moles of reactants.
Conclusion: Mastering Stoichiometry
Mastering stoichiometry is essential for anyone studying chemistry or related fields. It provides a fundamental understanding of chemical reactions and their quantitative aspects. By understanding the concepts of molar ratios, limiting reactants, and percentage yield, you can accurately predict and analyze the outcomes of chemical reactions. So keep practicing, and don't be afraid to tackle challenging problems. Happy chemistry!