Finding Oxygen Volume In Aluminum Oxide Formation

Hey guys! Let's dive into a cool chemistry problem. Imagine you're in a lab, and you've got some aluminum metal reacting with oxygen. This reaction creates aluminum oxide ($Al_2O_3$), and you've got some specific data about the experiment. Our mission? To figure out the volume of oxygen ($O_2$) that was used. It's like a real-life puzzle, and we'll break it down step by step to make sure it's super clear. Understanding this type of calculation is a cornerstone of chemistry, helping us predict and understand how much of a reactant is used, or a product is formed. Ready? Let's get started!

Understanding the Chemical Reaction

First off, let's look at the chemical equation. This is the recipe for our reaction:

$3 O_2 + 4 Al \rightarrow 2 Al_2O_3$

This equation tells us that three molecules of oxygen ($O_2$) react with four atoms of aluminum (Al) to produce two molecules of aluminum oxide ($Al_2O_3$). This equation is balanced, meaning that the number of atoms of each element is the same on both sides of the arrow. This balance is crucial because it follows the law of conservation of mass – that matter is neither created nor destroyed in a chemical reaction. It's all about keeping track of the atoms! Without a balanced equation, all calculations would be pointless.

The Importance of Stoichiometry

This also brings in stoichiometry, which is the study of the quantitative relationships or ratios between reactants and products in a chemical reaction. Stoichiometry is super important because it allows us to predict the amount of reactants needed or products formed in a reaction. Without it, chemists wouldn't be able to accurately perform any experiment! By using the balanced equation, we can determine the mole ratios that allow us to convert between the amounts of substances involved in the reaction. These mole ratios are the coefficients in front of each chemical formula in the balanced equation. We can calculate the amount of $O_2$ consumed during the experiment using stoichiometry with the mass of aluminum oxide.

The Given Information: What We Know

Here's what we know from our lab experiment:

• Mass of Aluminum Oxide ($Al_2O_3$): 46.54 grams
• Temperature (T): 300.0 K (Kelvin)
• Pressure (P): 1.2 atm (atmospheres)

We will use this info, along with some constants to get our answer! We will also need the molar mass of each compound.

Molar Masses

To make sure we get our units correct, we'll need the molar masses of the compounds involved. The molar mass is the mass of one mole of a substance. It's like saying, "how much does a dozen of these things weigh?" We can get the molar masses from the periodic table, so let's find them:

• Molar mass of $Al_2O_3$: 101.96 g/mol (approximately)
• Molar mass of $O_2$: 32.00 g/mol (approximately)

These molar masses will let us convert between grams and moles, which is essential for our calculations.

Calculating the Moles of Aluminum Oxide ($Al_2O_3$)

Now, let's figure out how many moles of $Al_2O_3$ were formed. We'll use the mass of $Al_2O_3$ (46.54 grams) and its molar mass (101.96 g/mol) to do this. Remember that moles are like a unit of measurement for the amount of a substance, similar to how we use grams or kilograms for mass.

Step-by-Step Calculation:

1. Use the formula: Moles = Mass / Molar Mass
2. Plug in the values: Moles of $Al_2O_3$ = 46.54 g / 101.96 g/mol
3. Calculate: Moles of $Al_2O_3$ ≈ 0.4565 moles

So, 0.4565 moles of aluminum oxide were produced in the experiment. This conversion from grams to moles is a critical step because it allows us to use the balanced chemical equation to relate the amount of $Al_2O_3$ formed to the amount of $O_2$ consumed.

Finding the Moles of Oxygen ($O_2$) Used

Alright, now that we know the moles of $Al_2O_3$ formed, we can use the balanced chemical equation to figure out how many moles of $O_2$ were used. This is where the stoichiometry we talked about earlier comes into play!

Using the Mole Ratio

1. Look at the balanced equation: $3 O_2 + 4 Al \rightarrow 2 Al_2O_3$
2. Find the mole ratio: The equation tells us that 3 moles of $O_2$ react to produce 2 moles of $Al_2O_3$. This means the mole ratio of $O_2$ to $Al_2O_3$ is 3:2.
3. Calculate the moles of $O_2$: Moles of $O_2$ = (Moles of $Al_2O_3$) × (Mole ratio of $O_2$ to $Al_2O_3$)
4. Plug in the values: Moles of $O_2$ = 0.4565 moles × (3 moles $O_2$ / 2 moles $Al_2O_3$)
5. Calculate: Moles of $O_2$ ≈ 0.6848 moles

So, approximately 0.6848 moles of oxygen were used in the experiment. This calculation is a key part of the problem. It highlights the importance of the balanced chemical equation in determining the quantitative relationships between reactants and products.

Calculating the Volume of Oxygen ($O_2$) Using the Ideal Gas Law

Now, we're on the last stretch! We have the moles of oxygen ($O_2$), and we know the temperature and pressure. We can use the Ideal Gas Law to find the volume of oxygen used.

The Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of gases. The formula is:

$PV = nRT$

Where:

• P = Pressure (in atm)
• V = Volume (in liters)
• n = Moles (in moles)
• R = Ideal gas constant (0.0821 L·atm/mol·K)
• T = Temperature (in Kelvin)

Plugging in the Values and Solving

1. Rearrange the formula to solve for V: $V = \frac{nRT}{P}$
2. Plug in the values: V = (0.6848 mol × 0.0821 L·atm/mol·K × 300.0 K) / 1.2 atm
3. Calculate: V ≈ 14.02 liters

Therefore, the volume of $O_2$ used during the experiment was approximately 14.02 liters. None of the answer choices is correct. It is important to review the problem and calculations again.

Reviewing the Calculations and Conclusion

Let's go over everything one more time. We started with the mass of $Al_2O_3$ formed, converted it to moles, used stoichiometry to find the moles of $O_2$ used, and then used the Ideal Gas Law to find the volume of $O_2$. Chemistry is all about problem-solving and understanding how matter behaves. By breaking down the problem step-by-step, we were able to calculate the volume of oxygen used in the reaction. Pretty cool, right?

Final Answer

The volume of $O_2$ used during the experiment is approximately 14.02 liters.