Aluminum Reaction With Nitric Acid: Gas Volume Calculation

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Aluminum Reaction with Nitric Acid: Gas Volume Calculation

Hey guys! Let's dive into a fascinating chemistry problem involving the reaction of aluminum with nitric acid. We're going to figure out how much nitrogen gas is produced when 54 grams of aluminum reacts completely. This is a classic stoichiometry problem, and I'll walk you through each step so it's super clear. So, grab your calculators, and let's get started!

Understanding the Chemical Reaction

First off, let's write down the unbalanced chemical equation:

AI + HNO3 → AI(NO3)3 + N₂ + H₂O

Before we can calculate anything, we need to balance this equation. Balancing ensures that we have the same number of atoms for each element on both sides of the equation, which is crucial for stoichiometry. Balancing chemical equations can sometimes feel like solving a puzzle, but there's a systematic way to do it. We'll start by counting the number of atoms for each element on both sides and then adjust the coefficients to make them equal.

Balancing the Equation: A Step-by-Step Approach

Let's break it down:

  1. Aluminum (Al): We have 1 Al atom on both sides, so that's balanced for now.
  2. Hydrogen (H): We have 1 H atom on the reactant side (HNO3) and 2 H atoms on the product side (Hâ‚‚O). We'll need to adjust this.
  3. Nitrogen (N): We have 1 N atom on the reactant side (HNO3) and multiple N atoms on the product side (AI(NO3)3 and Nâ‚‚). This looks like a key area to focus on.
  4. Oxygen (O): Oxygen atoms are present in HNO3, AI(NO3)3, and Hâ‚‚O, making it a bit trickier to balance directly. It's often a good idea to leave oxygen for later.

Now, let's start adjusting the coefficients. A common strategy is to balance metals first, then non-metals, and finally hydrogen and oxygen. Since aluminum is already balanced, we'll move on to nitrogen. The aluminum nitrate [AI(NO3)3] has 3 nitrogen atoms, and we also have Nâ‚‚ on the product side. This suggests we'll need to play around with the coefficients of HNO3 and Nâ‚‚. Balancing chemical equations is a bit like trial and error, but with practice, you'll get the hang of it! Start by making educated guesses and then check if the number of atoms balances out. If not, adjust the coefficients accordingly until you get the same number of each type of atom on both sides of the equation.

The Balanced Equation

After some balancing, we arrive at the balanced equation:

10 AI + 36 HNO3 → 10 AI(NO3)3 + 3 N₂ + 18 H₂O

Take a moment to verify this. Count the atoms on each side:

  • Aluminum (Al): 10 on both sides
  • Hydrogen (H): 36 on both sides
  • Nitrogen (N): 36 on both sides
  • Oxygen (O): 108 on both sides

Perfect! Everything balances. Now that we have a balanced equation, we can move on to the stoichiometry calculations.

Stoichiometry: Moles and Molar Mass

Next, we need to figure out how many moles of aluminum (Al) we have. Remember, the molar mass of a substance is the mass of one mole of that substance, usually expressed in grams per mole (g/mol). We're given that the molar mass of aluminum is 27 g/mol, and we have 54 grams of Al. To find the number of moles, we use the formula:

Moles = Mass / Molar Mass

So, for aluminum:

Moles of Al = 54 g / 27 g/mol = 2 moles

Now we know we have 2 moles of aluminum reacting. This is a crucial piece of information because it allows us to use the balanced equation to determine how many moles of nitrogen gas (Nâ‚‚) will be produced.

Mole Ratio from the Balanced Equation

The balanced equation tells us the mole ratio between the reactants and products. In our balanced equation:

10 AI + 36 HNO3 → 10 AI(NO3)3 + 3 N₂ + 18 H₂O

We see that 10 moles of Al react to produce 3 moles of Nâ‚‚. This is our key mole ratio. We can write this as a fraction:

(3 moles Nâ‚‚) / (10 moles Al)

We'll use this ratio to convert the moles of Al we have (2 moles) to moles of Nâ‚‚ produced. This conversion is the heart of stoichiometry, allowing us to predict the amounts of products formed from a given amount of reactants. By using the mole ratio from the balanced equation, we can confidently calculate the theoretical yield of nitrogen gas in this reaction. Stoichiometry is such a cool tool, isn't it?

Calculating Moles of Nitrogen Gas

Now, let's calculate the moles of Nâ‚‚ produced from 2 moles of Al. We'll use the mole ratio we just found:

Moles of Nâ‚‚ = (2 moles Al) * (3 moles Nâ‚‚ / 10 moles Al)

The "moles Al" units cancel out, leaving us with:

Moles of Nâ‚‚ = (2 * 3) / 10 = 0.6 moles

So, 2 moles of aluminum will produce 0.6 moles of nitrogen gas. We're getting closer to our final answer! Now we need to convert these moles of Nâ‚‚ to a volume under normal conditions.

Converting Moles to Volume: The Ideal Gas Law

To find the volume of Nâ‚‚ gas produced under normal conditions (also known as standard temperature and pressure, or STP), we'll use the ideal gas law. At STP, 1 mole of any ideal gas occupies 22.4 liters. This is a fundamental concept in chemistry and is super handy for these kinds of calculations.

The Magic Number: 22.4 Liters

Remember, the molar volume of a gas at STP is 22.4 liters per mole (L/mol). This means that if we have 1 mole of any gas (like nitrogen, oxygen, or even a mixture of gases behaving ideally), it will occupy a volume of 22.4 liters at 0°C (273.15 K) and 1 atmosphere of pressure. This constant is derived from the ideal gas law equation (PV = nRT), where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. At STP, these values are standardized, leading to the 22.4 L/mol relationship.

Calculating the Volume of Nitrogen Gas

We have 0.6 moles of Nâ‚‚ gas, and we know that 1 mole occupies 22.4 liters at STP. So, to find the volume, we multiply the number of moles by the molar volume:

Volume of Nâ‚‚ = (0.6 moles) * (22.4 L/mol)

Volume of Nâ‚‚ = 13.44 liters

Final Answer

Therefore, the volume of Nâ‚‚ gas produced under normal conditions when 54 grams of Al metal completely reacts is 13.44 liters.

So, there you have it! We've successfully solved this stoichiometry problem by balancing the equation, calculating moles, using mole ratios, and applying the ideal gas law. Chemistry can be challenging, but breaking it down step-by-step makes it much easier. Keep practicing, and you'll become a pro in no time! If you have any questions, feel free to ask. Happy chemistry-ing, everyone!