HCl Reacts With Aluminum Hydroxide: A Step-by-Step Guide

Hey chemistry enthusiasts! Today, we're diving into a classic stoichiometry problem: figuring out how much hydrochloric acid (HCl) is needed to react with a specific amount of aluminum hydroxide ($Al(OH)_3$). It's a fundamental concept, and understanding it unlocks a deeper understanding of chemical reactions. Let's break it down, step by step, so you can nail this kind of problem every time. We will explore the key concepts to determine the mass in grams of HCl that can react with 0.750 g of $Al(OH)_3$.

Understanding the Chemical Reaction

First things first, let's look at the balanced chemical equation. This equation is the recipe for our reaction, telling us exactly what's happening on a molecular level. The balanced equation for the reaction between aluminum hydroxide and hydrochloric acid is:

$Al(OH)_3(s) + 3 HCl(aq) \rightarrow AlCl_3(aq) + 3 H_2O(l)$

This equation tells us that one mole of solid aluminum hydroxide ($Al(OH)_3$) reacts with three moles of aqueous hydrochloric acid ($HCl$) to produce one mole of aluminum chloride ($AlCl_3$) and three moles of water ($H_2O$). The coefficients in front of each chemical formula (the numbers like 3 in front of HCl and $H_2O$) are super important; they represent the mole ratio, which is the heart of our calculations. Understanding the balanced equation is the cornerstone of solving stoichiometry problems. It gives us the mole ratios that we will use in our calculations. The coefficients are critical in understanding the quantitative relationships within the reaction. They are the foundation for converting between the amount of reactants and products in a chemical reaction. Without a balanced equation, we would be completely lost.

Now, let's get down to the specifics. We know we have 0.750 grams of $Al(OH)_3$ to start with. Our goal is to figure out how many grams of HCl are needed to react completely with that amount. This is where the magic of stoichiometry comes in! We need to convert the grams of $Al(OH)_3$ into moles, use the mole ratio from the balanced equation to find the moles of HCl needed, and then convert those moles of HCl back into grams. It sounds like a few steps, but I promise, it's not as complicated as it seems once you get the hang of it.

Step-by-Step Calculation: Finding the Mass of HCl

Alright, buckle up, because we're about to crunch some numbers! Here's a detailed breakdown of the process:

Step 1: Convert Grams of $Al(OH)_3$ to Moles

We begin by converting the mass of aluminum hydroxide into moles. To do this, we need the molar mass of $Al(OH)_3$. The molar mass is the mass of one mole of a substance. You can calculate it by adding up the atomic masses of all the atoms in the chemical formula. Using the periodic table, we find the atomic masses:

• Al (Aluminum): 26.98 g/mol
• O (Oxygen): 16.00 g/mol
• H (Hydrogen): 1.01 g/mol

So, the molar mass of $Al(OH)_3$ is:

26.98 g/mol (Al) + 3 * (16.00 g/mol (O) + 1.01 g/mol (H)) = 78.01 g/mol

Now, we can convert the 0.750 g of $Al(OH)_3$ to moles using the molar mass:

Moles of $Al(OH)_3$ = (0.750 g) / (78.01 g/mol) = 0.00961 mol

So, we have 0.00961 moles of $Al(OH)_3$ to work with. We have now successfully converted the mass of aluminum hydroxide to moles. This is an important first step in stoichiometry, as it allows us to relate the amount of one substance to another using the mole ratio from the balanced chemical equation. Remember, the mole is the central unit in stoichiometry, and everything revolves around it. The molar mass is the bridge between grams and moles, and it's a vital tool for all of your stoichiometric calculations. Mastering this conversion is key to confidently tackling more complex problems.

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

This is where the balanced equation comes into play. Looking back at the equation, $Al(OH)_3(s) + 3 HCl(aq) \rightarrow AlCl_3(aq) + 3 H_2O(l)$, we see that 1 mole of $Al(OH)_3$ reacts with 3 moles of HCl. This gives us the mole ratio:

3 moles HCl / 1 mole $Al(OH)_3$

Now, we can use this ratio to calculate the moles of HCl needed to react with 0.00961 moles of $Al(OH)_3$:

Moles of HCl = 0.00961 mol $Al(OH)_3$ * (3 mol HCl / 1 mol $Al(OH)_3$) = 0.0288 mol HCl

We've now found that 0.0288 moles of HCl are required to react with the given amount of aluminum hydroxide. The mole ratio is the heart of stoichiometry, and it's essential to correctly set up this conversion. Always make sure that the units cancel out correctly, leaving you with the desired unit. The mole ratio allows us to convert from one substance to another. It's the secret sauce for chemical calculations! Practice setting up the ratios, and you'll become a master of stoichiometry in no time.

Step 3: Convert Moles of HCl to Grams

Our final step is to convert the moles of HCl into grams. To do this, we need the molar mass of HCl. Using the periodic table:

• H (Hydrogen): 1.01 g/mol
• Cl (Chlorine): 35.45 g/mol

The molar mass of HCl is 1.01 g/mol + 35.45 g/mol = 36.46 g/mol.

Now, we can convert moles of HCl to grams:

Mass of HCl = 0.0288 mol * (36.46 g/mol) = 1.05 g

Therefore, 1.05 grams of HCl are needed to react completely with 0.750 g of $Al(OH)_3$. We've successfully calculated the mass of HCl required. Congratulations! This step involves using the molar mass as a conversion factor once again, this time to go from moles to grams. Ensure that you use the correct molar mass and that your units cancel out properly to get the desired result. This final step completes the stoichiometry problem, linking the initial mass of the reactant to the mass of the other reactant. Remember to always review your work and consider whether your answer makes sense in the context of the problem. A bit of careful checking can catch any errors before you submit your answer.

Conclusion: Key Takeaways

So, what have we learned, guys? We've learned how to approach a stoichiometry problem step-by-step:

1. Start with a balanced chemical equation. This is your roadmap.
2. Convert grams to moles using molar mass.
3. Use the mole ratio from the balanced equation to find the moles of the unknown substance.
4. Convert moles back to grams using the molar mass.

By following these steps, you can confidently tackle any stoichiometry problem. Remember to pay close attention to the units and make sure they cancel out correctly. Practice makes perfect, so keep working through problems, and you'll become a stoichiometry pro in no time! Stoichiometry might seem daunting at first, but with practice and a systematic approach, it becomes a manageable and even enjoyable part of chemistry. Keep practicing, and you'll be solving these problems with ease. Stoichiometry is a foundational skill in chemistry, and mastering it will open doors to a deeper understanding of chemical reactions. Now, go forth and conquer those chemical equations!