Grams Of Br2 Required To React With AlCl3 Stoichiometry Problem

Hey guys! Today, we're diving into a stoichiometry problem that's going to help us figure out how much bromine ($Br_2$) is needed to react completely with 36.2 grams of aluminum chloride ($AlCl_3$). We've got a balanced chemical equation to guide us, so let's break it down step by step. If you've ever felt lost in the world of chemical reactions and molar masses, don't worry, we're going to make it super clear and easy to follow. Let's get started and solve this awesome chemistry puzzle together!

The Balanced Chemical Equation

Before we jump into any calculations, let's take a close look at the balanced chemical equation we're working with:

$$2 AlCl_3 + 3 Br_2 \rightarrow 2 AlBr_3 + 3 Cl_2$$

This equation is the key to understanding the relationship between the reactants and products in our chemical reaction. What it tells us is that two moles of aluminum chloride ($AlCl_3$) react with three moles of bromine ($Br_2$) to produce two moles of aluminum bromide ($AlBr_3$) and three moles of chlorine ($Cl_2$). This mole ratio is super important because it's the foundation for all our calculations. Think of it as a recipe: to make a certain amount of product, you need specific amounts of ingredients, and in our case, those ingredients are the reactants.

The coefficients in front of each chemical formula represent the number of moles involved in the reaction. So, for every 2 moles of $AlCl_3$ that react, we need 3 moles of $Br_2$. This 2:3 ratio is crucial. If we have a different amount of $AlCl_3$, we'll need to adjust the amount of $Br_2$ accordingly to ensure the reaction goes to completion. This is the essence of stoichiometry – using these ratios to figure out the exact quantities of reactants and products involved in a chemical reaction. Without a balanced equation, we'd be flying blind, so always make sure your equation is balanced before proceeding with any calculations. It’s like having the correct recipe before you start baking; otherwise, you might end up with a chemistry catastrophe!

Step 1: Convert Grams of AlCl₃ to Moles

Our first step is to convert the given mass of aluminum chloride ($AlCl_3$) into moles. Why moles? Because moles are the language of chemistry! The balanced equation speaks in terms of moles, so we need to convert our grams into moles to use the mole ratio effectively. To do this, we need the molar mass of $AlCl_3$.

The molar mass is the mass of one mole of a substance, and we can calculate it by adding up the atomic masses of each element in the compound from the periodic table. For $AlCl_3$, we have one aluminum atom (Al) and three chlorine atoms (Cl).

• The atomic mass of Aluminum (Al) is approximately 26.98 g/mol.
• The atomic mass of Chlorine (Cl) is approximately 35.45 g/mol.

So, the molar mass of $AlCl_3$ is:

$$Molar\ Mass\ (AlCl_3) = 26.98\ g/mol + 3 imes 35.45\ g/mol = 26.98\ g/mol + 106.35\ g/mol = 133.33\ g/mol$$

Now that we have the molar mass, we can convert the given mass of $AlCl_3$ (36.2 grams) to moles using the formula:

$$Moles = \frac{Mass}{Molar\ Mass}$$

Plugging in the values:

$$Moles\ (AlCl_3) = \frac{36.2\ g}{133.33\ g/mol} \approx 0.271\ mol$$

So, we have approximately 0.271 moles of $AlCl_3$. This is a crucial intermediate value because it connects the mass we were given to the mole ratio in our balanced equation. Think of it like translating a sentence from one language to another – we've just translated grams into moles, which is a language the balanced equation understands perfectly. Now we're ready to use this information to figure out how many moles of $Br_2$ we need.

Step 2: Use the Mole Ratio to Find Moles of Br₂

Okay, now that we know how many moles of $AlCl_3$ we have, we can use the mole ratio from the balanced equation to figure out how many moles of $Br_2$ are needed. Remember, the balanced equation tells us that 2 moles of $AlCl_3$ react with 3 moles of $Br_2$. This gives us a mole ratio of 3 moles $Br_2$ / 2 moles $AlCl_3$.

To find the moles of $Br_2$ required, we'll multiply the moles of $AlCl_3$ we calculated in the previous step (0.271 mol) by this mole ratio:

$$Moles\ (Br_2) = Moles\ (AlCl_3) imes \frac{3\ mol\ Br_2}{2\ mol\ AlCl_3}$$

$$Moles\ (Br_2) = 0.271\ mol\ AlCl_3 imes \frac{3\ mol\ Br_2}{2\ mol\ AlCl_3}$$

$$Moles\ (Br_2) \approx 0.4065\ mol$$

So, we need approximately 0.4065 moles of $Br_2$ to react completely with 0.271 moles of $AlCl_3$. This step is like using a recipe ratio to scale up or down the ingredients. If you know you need twice the amount of cake, you double all the ingredients – similarly, we're using the mole ratio to find the exact amount of $Br_2$ needed for our specific amount of $AlCl_3$. We're almost there! We've got the moles of $Br_2$, and now we just need to convert it back into grams, which is what the question is asking for.

Step 3: Convert Moles of Br₂ to Grams

We're in the home stretch now! We know how many moles of $Br_2$ we need (0.4065 moles), and we need to convert that back into grams to answer the question. To do this, we'll use the molar mass of $Br_2$. Remember, the molar mass is the mass of one mole of a substance, and we can find it by looking up the atomic masses on the periodic table.

Bromine ($Br_2$) is a diatomic molecule, meaning it exists as two bromine atoms bonded together. The atomic mass of a single bromine (Br) atom is approximately 79.90 g/mol. So, the molar mass of $Br_2$ is:

$$Molar\ Mass\ (Br_2) = 2 imes 79.90\ g/mol = 159.80\ g/mol$$

Now we can convert moles of $Br_2$ to grams using the formula:

$$Mass = Moles imes Molar\ Mass$$

Plugging in the values:

$$Mass\ (Br_2) = 0.4065\ mol imes 159.80\ g/mol$$

$$Mass\ (Br_2) \approx 65.0\ grams$$

So, approximately 65.0 grams of $Br_2$ are required to react completely with 36.2 grams of $AlCl_3$. We've successfully navigated through the stoichiometry, converted grams to moles, used the mole ratio, and converted back to grams. It's like we've built a bridge from the given information to the answer we needed! Pat yourself on the back – you've tackled a significant chemistry problem.

Final Answer

Based on our calculations, the correct answer is approximately 65.0 grams of $Br_2$, which corresponds to answer choice A. 65.2 grams. Yay, we nailed it! This problem really highlights how important it is to understand the balanced equation and molar masses in stoichiometry. You've seen how we can take a seemingly complex question and break it down into manageable steps. Keep practicing, and you'll become a stoichiometry superstar in no time! Remember, chemistry is all about building understanding step by step, and you've just taken a giant leap forward. Awesome job, guys!