Calculating Gas Volume From Aluminum Dissolution In HCl

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Hey guys! Let's dive into a fascinating chemistry problem today. We're going to figure out how to calculate the volume of gas released when aluminum dissolves in hydrochloric acid. This is a classic stoichiometry problem that combines concepts of molar mass, concentration, and the ideal gas law. So, buckle up, and let's get started!

Understanding the Problem

First, let's break down the problem. We have a piece of aluminum reacting with a hydrochloric acid (HCl) solution. This reaction produces hydrogen gas (H₂) and aluminum chloride (AlCl₃). The key here is to find out how much hydrogen gas is produced, and then calculate its volume at standard conditions (STP). We're given the mass of the HCl solution (200 g) and the mass percentage of HCl in the solution (25%). This information is crucial for determining the amount of HCl actually reacting.

The main challenge in this type of problem is bridging the gap between the mass of the solution and the volume of gas produced. We'll need to use stoichiometry, which is essentially the math of chemical reactions, to connect these quantities. Remember, stoichiometry relies on the balanced chemical equation, which tells us the molar ratios of reactants and products. So, let's start by writing the balanced equation for this reaction.

Why is this important in real life? Well, understanding these calculations is vital in many industrial processes, especially in chemical manufacturing, where controlling gas production is essential for safety and efficiency. Plus, it’s just plain cool to see how chemistry works in action!

1. The Balanced Chemical Equation

The first and most crucial step is to write the balanced chemical equation for the reaction. This equation tells us the exact ratio in which the reactants combine and the products are formed. The unbalanced equation is:

Al + HCl → AlCl₃ + H₂

Now, let's balance it. We need to make sure that the number of atoms of each element is the same on both sides of the equation. Here's the balanced equation:

2Al + 6HCl → 2AlCl₃ + 3H₂

This balanced equation is super important. It tells us that 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 2 moles of aluminum chloride and 3 moles of hydrogen gas. This 2:6:2:3 ratio is the foundation of our stoichiometric calculations. If you mess this part up, the rest of your calculation will be incorrect, so double-check it!

Think of it like a recipe: The balanced equation is like a recipe that tells you exactly how much of each ingredient (reactants) you need to make a certain amount of the final product. In this case, we want to know how much hydrogen gas we're “cooking up.”

2. Calculating the Mass of HCl

Next, we need to figure out how much actual HCl is present in the 200 g solution. We know that the solution is 25% HCl by mass. This means that 25% of the 200 g solution is pure HCl. Let's calculate that:

Mass of HCl = (25/100) * 200 g = 50 g

So, we have 50 grams of hydrochloric acid in the solution. This is a critical piece of information because we need to convert this mass into moles to use the stoichiometric ratios from the balanced equation. Remember, stoichiometry works with moles, not grams, because moles represent the actual number of molecules involved in the reaction.

Imagine this: If you were baking a cake, you wouldn't measure flour in terms of the weight of the bag, you'd measure it in cups or grams, a unit that directly relates to the amount you need for the recipe. Similarly, in chemistry, moles are the units that directly relate to the number of molecules reacting.

3. Converting Mass of HCl to Moles

To convert the mass of HCl to moles, we need to use its molar mass. The molar mass of HCl is the sum of the atomic masses of hydrogen (H) and chlorine (Cl):

Molar mass of H ≈ 1 g/mol Molar mass of Cl ≈ 35.5 g/mol Molar mass of HCl = 1 g/mol + 35.5 g/mol = 36.5 g/mol

Now we can convert the mass of HCl (50 g) to moles:

Moles of HCl = Mass of HCl / Molar mass of HCl Moles of HCl = 50 g / 36.5 g/mol ≈ 1.37 moles

We now know that approximately 1.37 moles of HCl are reacting in this solution. This is a key number because we'll use it with the balanced equation to find out how many moles of hydrogen gas are produced.

Think of molar mass as a conversion factor: It allows us to switch between grams (a macroscopic measurement we can easily make in the lab) and moles (a microscopic measurement that relates directly to the number of molecules).

4. Using Stoichiometry to Find Moles of H₂

This is where the balanced chemical equation really shines. We know from the equation:

2Al + 6HCl → 2AlCl₃ + 3H₂

that 6 moles of HCl produce 3 moles of H₂. This gives us a mole ratio of 3 moles H₂ / 6 moles HCl, which simplifies to 1 mole H₂ / 2 moles HCl. We can use this ratio to find the moles of H₂ produced:

Moles of H₂ = Moles of HCl * (Moles of H₂ / Moles of HCl) Moles of H₂ = 1.37 moles HCl * (3 moles H₂ / 6 moles HCl) Moles of H₂ = 1.37 moles HCl * (1/2) Moles of H₂ ≈ 0.685 moles

So, approximately 0.685 moles of hydrogen gas are produced in this reaction. We're getting closer to our final answer! This step highlights the power of stoichiometry – it allows us to predict the amount of product formed based on the amount of reactant we start with.

Remember, ratios are your friends: In stoichiometry, mole ratios are the tools that allow you to move from one substance to another in a chemical reaction. They are derived directly from the balanced chemical equation.

5. Calculating the Volume of H₂ at STP

Now that we know the moles of H₂ produced, we can calculate its volume at standard temperature and pressure (STP). STP is defined as 0°C (273.15 K) and 1 atmosphere (atm) pressure. At STP, 1 mole of any ideal gas occupies 22.4 liters. This is a crucial constant to remember!

We can use this molar volume to find the volume of 0.685 moles of H₂:

Volume of H₂ = Moles of H₂ * Molar volume at STP Volume of H₂ = 0.685 moles * 22.4 L/mol Volume of H₂ ≈ 15.34 L

Therefore, approximately 15.34 liters of hydrogen gas are released when the aluminum dissolves in the hydrochloric acid solution at STP. That's our final answer! We've successfully navigated the stoichiometric calculations and arrived at the volume of gas produced.

Why STP matters: STP provides a standard reference point for comparing gas volumes. Because gases are highly compressible, their volume changes significantly with temperature and pressure. Specifying STP allows us to compare gas volumes under the same conditions.

Summary of Steps

Let's recap the steps we took to solve this problem:

  1. Balanced the chemical equation: 2Al + 6HCl → 2AlCl₃ + 3H₂
  2. Calculated the mass of HCl: 50 g
  3. Converted mass of HCl to moles: ≈ 1.37 moles
  4. Used stoichiometry to find moles of H₂: ≈ 0.685 moles
  5. Calculated the volume of H₂ at STP: ≈ 15.34 L

By following these steps, you can solve similar stoichiometry problems involving gas volumes. Remember, the key is to understand the balanced equation, convert masses to moles, use mole ratios, and apply the ideal gas law (or the molar volume at STP).

Common Mistakes to Avoid

  • Forgetting to balance the equation: This is the cardinal sin of stoichiometry! An unbalanced equation will lead to incorrect mole ratios and a wrong answer.
  • Using grams directly in stoichiometric calculations: You must convert grams to moles before using mole ratios from the balanced equation.
  • Using the wrong molar mass: Double-check that you're using the correct molar mass for each compound.
  • Mixing up units: Make sure you're consistent with your units throughout the calculation.
  • Not understanding STP: Remember that 1 mole of any ideal gas occupies 22.4 L at STP. This is a crucial constant.

Practice Makes Perfect

Stoichiometry can seem daunting at first, but with practice, it becomes second nature. Try solving similar problems with different reactants and products. The more you practice, the more confident you'll become in your ability to tackle these types of calculations.

So, there you have it, guys! We've successfully calculated the volume of gas released during a chemical reaction. I hope this step-by-step guide has been helpful. Keep practicing, and you'll be a stoichiometry master in no time! Now go and impress your chemistry teacher! 😉