Ferrous Chloride Yield: A Stoichiometry Calculation Guide

Hey everyone! Let's dive into a chemistry problem together. We're going to figure out how much ferrous chloride (FeCl₂) we can get from reacting iron (Fe) with hydrochloric acid (HCl). This is a classic stoichiometry problem, and we'll break it down step by step. We'll start with the basics, go through the calculations, and even consider the reaction efficiency. So, buckle up, and let's get started!

The Problem: Iron and Hydrochloric Acid

Okay, so here’s the deal. We're reacting 11 grams of iron with 0.3 moles of hydrochloric acid. The big question is: how many grams of ferrous chloride will we collect if the reaction is only 92% efficient? This 92% efficiency part is crucial because, in the real world, reactions rarely go perfectly. There's always some loss due to side reactions or incomplete conversions. This is important because it tells us we can't just calculate the theoretical yield and call it a day. We need to factor in this efficiency to get a realistic answer. So, let's break down each part of the problem and solve it. Our focus will be on understanding the reaction, the stoichiometry involved, and how to handle that pesky percentage yield. We'll go through each step meticulously, ensuring that every calculation is clear and understandable. This will help us not only solve this particular problem but also equip us with the skills to tackle similar challenges in the future. Remember, chemistry is all about understanding the relationships between different substances and how they interact, so let’s get our hands dirty and dive into the nitty-gritty details of this reaction!

Step 1: The Balanced Chemical Equation

First things first, we need the balanced chemical equation. This is the foundation of any stoichiometry problem. It tells us exactly how many moles of each reactant are needed to produce a certain number of moles of product. For this reaction, iron (Fe) reacts with hydrochloric acid (HCl) to produce ferrous chloride (FeCl₂) and hydrogen gas (H₂). The unbalanced equation looks like this:

Fe + HCl → FeCl₂ + H₂

Now, we need to balance it. We can see that there are two chlorine atoms on the product side (FeCl₂) but only one on the reactant side (HCl). To fix this, we'll put a coefficient of 2 in front of HCl. This also gives us two hydrogen atoms on both sides, so the equation is now balanced:

Fe + 2 HCl → FeCl₂ + H₂

This balanced equation is super important, guys, because it gives us the molar ratios we need for our calculations. It tells us that 1 mole of iron reacts with 2 moles of hydrochloric acid to produce 1 mole of ferrous chloride and 1 mole of hydrogen gas. This is the key to unlocking the rest of the problem. Without a balanced equation, our calculations would be meaningless. It’s like trying to build a house without a blueprint – you might end up with something, but it probably won't be what you intended. So, always make sure your equation is balanced before moving on. This simple step can save you a lot of headaches down the road. Now that we have our balanced equation, we can confidently move on to the next step: figuring out the molar masses of our reactants and products.

Step 2: Calculate Molar Masses

Next up, we need to calculate the molar masses of iron (Fe) and ferrous chloride (FeCl₂). The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). We can find these values on the periodic table. For iron (Fe), the molar mass is approximately 55.85 g/mol. This means that one mole of iron weighs 55.85 grams. For ferrous chloride (FeCl₂), we need to add the molar mass of iron to twice the molar mass of chlorine. The molar mass of chlorine (Cl) is approximately 35.45 g/mol. So, the molar mass of FeCl₂ is:

55. 85 g/mol (Fe) + 2 * 35.45 g/mol (Cl) = 126.75 g/mol

So, one mole of ferrous chloride weighs 126.75 grams. These molar masses are essential for converting between grams and moles, which is a crucial step in stoichiometry problems. Think of molar mass as the bridge that connects the mass world (grams) to the mole world (the number of particles). We're given the mass of iron in grams, but the balanced equation deals with moles. So, we need to convert grams of iron to moles of iron. Similarly, we'll calculate the moles of ferrous chloride from the balanced equation, and then use the molar mass of ferrous chloride to convert back to grams. This conversion process is at the heart of stoichiometric calculations. Understanding molar masses and how to use them is a fundamental skill in chemistry. Without this knowledge, we wouldn't be able to accurately predict the amount of product formed in a reaction. Now that we've got our molar masses sorted out, we're well-equipped to tackle the next step: converting the given mass of iron into moles.

Step 3: Convert Grams of Iron to Moles

Now, let's convert the given mass of iron (11 grams) into moles. We'll use the molar mass of iron that we calculated in the previous step (55.85 g/mol). To do this, we'll divide the mass of iron by its molar mass:

Moles of Fe = Mass of Fe / Molar mass of Fe Moles of Fe = 11 g / 55.85 g/mol Moles of Fe ≈ 0.197 moles

So, we have approximately 0.197 moles of iron. This conversion is a pivotal step because it allows us to compare the amount of iron with the amount of hydrochloric acid in terms of moles, which is what the balanced equation uses. Remember, the balanced equation tells us the ratio in which substances react in terms of moles, not grams. Converting to moles allows us to use these ratios directly. Think of it like this: grams are like different currencies, and moles are like a universal exchange rate. We need to convert everything into the same 'currency' (moles) before we can compare and calculate. This is why this step is so crucial. It’s the bridge that allows us to go from the macroscopic world (grams, what we can measure in the lab) to the microscopic world (moles, the number of particles interacting in the reaction). Now that we know the moles of iron, we're ready to figure out the limiting reactant. This will tell us which reactant is going to run out first and, therefore, dictate how much product we can make. So, let's move on to the next step and identify our limiting reactant!

Step 4: Determine the Limiting Reactant

To figure out the limiting reactant, we need to compare the moles of iron (0.197 moles) with the moles of hydrochloric acid (0.3 moles) given in the problem. Remember the balanced equation: Fe + 2 HCl → FeCl₂ + H₂. This tells us that 1 mole of iron reacts with 2 moles of hydrochloric acid. Now, let's see how much HCl we would need to react completely with 0.197 moles of iron:

Moles of HCl needed = Moles of Fe * (2 moles HCl / 1 mole Fe) Moles of HCl needed = 0.197 moles * 2 Moles of HCl needed ≈ 0.394 moles

We need 0.394 moles of HCl to react completely with all the iron. But we only have 0.3 moles of HCl. This means that HCl will run out before all the iron can react. Therefore, hydrochloric acid (HCl) is the limiting reactant. The limiting reactant is super important because it dictates the maximum amount of product that can be formed. It’s like having a recipe that calls for two eggs per cup of flour, but you only have one egg. You can’t use all the flour, can you? The egg is your limiting ingredient, and it limits how much of the final product you can make. Similarly, in this reaction, the amount of ferrous chloride we can produce is limited by the amount of hydrochloric acid we have. The iron is in excess, meaning we have more than enough of it to react with the available HCl. So, from now on, we'll base our calculations on the amount of the limiting reactant, which is HCl. Knowing the limiting reactant is a crucial step in accurately predicting the yield of a reaction. Now that we've identified HCl as our limiting reactant, we can confidently move on to the next step: calculating the theoretical yield of ferrous chloride.

Step 5: Calculate the Theoretical Yield of Ferrous Chloride

The theoretical yield is the maximum amount of product that can be formed if the reaction goes to completion, assuming no losses. We'll use the moles of the limiting reactant (HCl) and the balanced equation to calculate this. The balanced equation tells us that 2 moles of HCl produce 1 mole of FeCl₂. We have 0.3 moles of HCl, so let's calculate how many moles of FeCl₂ we can produce:

Moles of FeCl₂ = Moles of HCl * (1 mole FeCl₂ / 2 moles HCl) Moles of FeCl₂ = 0.3 moles * (1/2) Moles of FeCl₂ = 0.15 moles

So, theoretically, we can produce 0.15 moles of ferrous chloride. Now, we need to convert this to grams using the molar mass of FeCl₂ (126.75 g/mol):

Mass of FeCl₂ = Moles of FeCl₂ * Molar mass of FeCl₂ Mass of FeCl₂ = 0.15 moles * 126.75 g/mol Mass of FeCl₂ ≈ 19.01 grams

Therefore, the theoretical yield of ferrous chloride is approximately 19.01 grams. This is the amount of product we would expect if everything went perfectly, and there were no losses. But, as we discussed earlier, reactions rarely go perfectly in the real world. There are always some losses due to various factors, such as incomplete reactions or side reactions. That's where the concept of percent yield comes in. The theoretical yield is a great starting point, but the actual yield (the amount of product we actually obtain) is usually less. This difference is expressed as the percent yield. So, now that we have our theoretical yield, we're ready to factor in the given efficiency of the reaction (92%) and calculate the actual yield. This will give us a more realistic estimate of how much ferrous chloride we'll actually collect. Let’s move on to the final step and put it all together!

Step 6: Calculate the Actual Yield Considering the Efficiency

Finally, we need to consider the efficiency of the reaction, which is given as 92%. This means that we only get 92% of the theoretical yield in reality. To calculate the actual yield, we'll multiply the theoretical yield (19.01 grams) by the efficiency:

Actual yield of FeCl₂ = Theoretical yield * Efficiency Actual yield of FeCl₂ = 19.01 grams * 0.92 Actual yield of FeCl₂ ≈ 17.49 grams

So, considering the 92% efficiency, we can expect to collect approximately 17.49 grams of ferrous chloride. This is our final answer! This step is crucial because it bridges the gap between the ideal world of theoretical calculations and the real world of laboratory experiments. The percent yield is a way of quantifying how 'successful' a reaction was. A higher percent yield means that more of the reactants were converted into the desired product, with minimal losses due to side reactions or other factors. A lower percent yield, on the other hand, indicates that the reaction was less efficient, and more product was lost along the way. Understanding percent yield is not just about getting the right answer in a problem; it's about understanding the practical limitations of chemical reactions and how to optimize them. In a real-world setting, chemists are constantly striving to improve reaction efficiencies to minimize waste and maximize the production of desired products. And there we have it, guys! We've successfully navigated through this stoichiometry problem, step by step, from balancing the equation to calculating the actual yield. We've seen how each step builds upon the previous one, and how important it is to have a solid understanding of the underlying concepts. Now, let’s recap the key takeaways from this exercise.

Key Takeaways

Let's recap what we've learned in this stoichiometry adventure! First, balancing the chemical equation is fundamental. It provides the crucial molar ratios needed for calculations. Without a balanced equation, the entire calculation falls apart. Think of it as the grammar of chemistry – it ensures that your 'sentences' (reactions) make sense. Next, calculating molar masses allows us to convert between grams and moles, which is essential because the balanced equation speaks in terms of moles. Molar mass is our conversion factor, the bridge between the macroscopic world (grams) and the microscopic world (moles). Then, converting grams of reactants to moles allows us to use the molar ratios from the balanced equation. It's like translating from one language to another so we can understand the relationship between the reactants. Determining the limiting reactant is vital because it dictates the maximum amount of product that can be formed. It's like identifying the weakest link in a chain – it determines the chain's overall strength. Calculating the theoretical yield gives us the ideal amount of product we could obtain if the reaction went perfectly. It's our benchmark, our target to aim for. Finally, considering the efficiency of the reaction and calculating the actual yield provides a realistic estimate of the product we'll collect. This is where we bring in the real-world constraints and acknowledge that reactions are not always perfect. By mastering these steps, you can tackle a wide range of stoichiometry problems with confidence. Remember, practice makes perfect! The more you work through these types of problems, the more comfortable you'll become with the concepts and calculations. And always remember to double-check your work, especially the balanced equation and the molar masses. A small error in the beginning can propagate through the entire calculation and lead to a wrong answer. Keep practicing, keep exploring, and keep enjoying the fascinating world of chemistry!