How To Calculate Grams Of Oxygen Needed For Reaction With Aluminum

Hey guys! Let's dive into a fun chemistry problem today. We're going to figure out just how much oxygen gas ($O_2$) we need to completely react with 9.30 moles of aluminum (Al). We've got a balanced chemical equation to help us out:

$4 Al + 3 O_2 \rightarrow 2 Al_2 O_3$

This equation tells us that 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide ($Al_2 O_3$). So, let's break down the problem step-by-step and make sure we really nail this concept. We'll go through what's given in the problem, how to use the stoichiometry, and finally, how to calculate the grams of oxygen needed. Ready? Let's get started!

What's Given in the Problem?

Okay, first things first, we need to figure out what information the problem is giving us. Identifying the 'givens' is super important in any chemistry problem. It’s like finding the ingredients before you start baking a cake! In this case, the problem clearly states that we have 9.30 moles of aluminum (Al). That's our starting point. We also have the balanced chemical equation:

$4 Al + 3 O_2 \rightarrow 2 Al_2 O_3$

This equation is super crucial because it tells us the molar ratio between aluminum and oxygen. Think of it as the recipe for our chemical reaction. For every 4 moles of aluminum, we need 3 moles of oxygen gas. This ratio is the key to solving the problem. So, to recap, the givens are:

• 9.30 moles of Aluminum (Al)
• The balanced chemical equation: $4 Al + 3 O_2 \rightarrow 2 Al_2 O_3$

Now that we know what we're starting with, we can move on to figuring out what the problem is asking us to find. In this case, we need to find out how many grams of oxygen gas ($O_2$) are required. So, we're going to use the information we have to convert moles of aluminum into grams of oxygen. Sounds like a plan, right? Let’s move on to the next step and see how we can use stoichiometry to solve this problem. Trust me, it's not as scary as it sounds!

Moles of Aluminum: The Starting Point

The problem explicitly tells us that we have 9.30 moles of aluminum. This is our initial value, the cornerstone upon which we will build our calculation. Think of it as the first domino in a series that will eventually lead us to the answer. Identifying this given is crucial because it sets the stage for the entire problem-solving process. Without knowing the initial quantity of aluminum, we wouldn't be able to determine how much oxygen is needed for the reaction.

The Balanced Chemical Equation: Our Reaction Recipe

The equation $4 Al + 3 O_2 \rightarrow 2 Al_2 O_3$ is far more than just a symbolic representation of a chemical reaction; it’s a detailed recipe that specifies the exact proportions in which the reactants combine. This equation tells us that for every 4 moles of aluminum (Al) that react, 3 moles of oxygen gas ($O_2$) are required. This relationship, known as the stoichiometric ratio, is the key to unlocking the solution. It's like knowing that a cake recipe calls for 2 cups of flour for every 1 cup of sugar – change the ratio, and you change the outcome. Understanding this ratio allows us to convert moles of aluminum to moles of oxygen, which is a critical step in our calculation.

Identifying the Target: Grams of Oxygen Gas

The problem clearly asks us to find the grams of oxygen gas ($O_2$) needed to react completely with the given amount of aluminum. This is our ultimate goal. Knowing what we need to find is just as important as knowing what we're given. It provides direction and context for our calculations. We're not just trying to find any number; we're looking for a specific quantity – the mass of oxygen gas. This understanding helps us choose the correct steps and conversions needed to arrive at the final answer.

So, to summarize, the 'givens' in this problem are not just numbers or formulas; they are the essential pieces of information that guide us through the problem-solving process. We have the initial quantity of aluminum (9.30 moles), the balanced chemical equation (our reaction recipe), and a clear target (grams of oxygen gas). With these pieces in place, we are well-equipped to tackle the next steps in solving this problem. Let's move on and explore how we can use stoichiometry to convert moles of aluminum to moles of oxygen – the next crucial step in our chemical journey!

Using Stoichiometry to Find Moles of Oxygen

Alright, now that we've identified what we know, let's get into the nitty-gritty of how we use stoichiometry to solve this. Stoichiometry, in simple terms, is like using the ratios in our balanced equation as a conversion factor. Remember that balanced equation?

$4 Al + 3 O_2 \rightarrow 2 Al_2 O_3$

It tells us that 4 moles of aluminum (Al) react with 3 moles of oxygen ($O_2$). This is our golden ratio! We can write this as a fraction to use as a conversion factor. We want to find moles of $O_2$, so we'll put that on top:

$(3 ext{ mol } O_2) / (4 ext{ mol } Al)$

Now, we can use this conversion factor to go from moles of aluminum to moles of oxygen. We start with our given, 9.30 moles of aluminum, and multiply it by our conversion factor:

$9.30 ext{ mol } Al imes (3 ext{ mol } O_2) / (4 ext{ mol } Al)$

Notice how the 'mol Al' cancels out, leaving us with moles of $O_2$. If you do the math, you'll find that:

$9.30 imes (3/4) = 6.975 ext{ mol } O_2$

So, we need 6.975 moles of oxygen gas to react completely with 9.30 moles of aluminum. We're not done yet, though! The question asks for grams of oxygen, not moles. But don't worry, we're on the right track. Now we just need to convert moles to grams, and we'll do that using the molar mass of oxygen. Let's get into it!

The Mole Ratio: Our Stoichiometric Bridge

The mole ratio derived from the balanced chemical equation is the linchpin of stoichiometric calculations. In our case, the ratio $3 ext{ mol } O_2 : 4 ext{ mol } Al$ acts as a bridge, allowing us to convert the amount of aluminum we have (in moles) to the corresponding amount of oxygen required (also in moles). This ratio is not just a number; it's a fundamental relationship dictated by the law of conservation of mass. It ensures that the reaction occurs in the correct proportions, with neither reactant being completely used up before the other.

Setting Up the Conversion: Dimensional Analysis in Action

The beauty of stoichiometry lies in its systematic approach to problem-solving. By setting up the conversion using dimensional analysis, we ensure that our units cancel out correctly, leading us to the desired unit. This method is like following a well-marked trail; each step logically connects to the next. We start with our given quantity (9.30 mol Al), and we multiply it by the mole ratio, making sure that the unit we want to eliminate (mol Al) appears in the denominator and the unit we want to obtain (mol $O_2$) appears in the numerator. This ensures that the units cancel out correctly, leaving us with the desired unit.

Calculating Moles of Oxygen: The Result of the Stoichiometric Dance

Performing the calculation $9.30 ext{ mol } Al imes (3 ext{ mol } O_2) / (4 ext{ mol } Al) = 6.975 ext{ mol } O_2$ is the culmination of our stoichiometric reasoning. This result tells us that 6.975 moles of oxygen gas are required to react completely with 9.30 moles of aluminum. This is a crucial intermediate step; we've now quantified the amount of oxygen needed in moles. However, the problem asks for the answer in grams, so we're not quite finished yet. We've successfully navigated the mole ratio conversion, and we're now ready to move on to the final step: converting moles of oxygen to grams of oxygen. This involves using the molar mass of oxygen, a concept that links the microscopic world of moles to the macroscopic world of grams – the world we can measure in the lab.

So, we've seen how stoichiometry, with its reliance on mole ratios and dimensional analysis, provides a powerful tool for solving chemical problems. We've successfully converted moles of aluminum to moles of oxygen, and we're now poised to tackle the final conversion to grams. Let's keep going and bring this problem home!

Converting Moles of Oxygen to Grams

Okay, we're in the home stretch now! We know we need 6.975 moles of oxygen gas ($O_2$). The problem, however, asks for the answer in grams. So, we need to convert moles to grams, and for that, we need the molar mass of oxygen gas. Remember, the molar mass is the mass of one mole of a substance, and it's usually given in grams per mole (g/mol).

Oxygen gas ($O_2$) is a diatomic molecule, meaning it consists of two oxygen atoms. The atomic mass of a single oxygen atom is approximately 16.00 g/mol. So, the molar mass of $O_2$ is:

$2 imes 16.00 ext{ g/mol} = 32.00 ext{ g/mol}$

Now we have our conversion factor! We know that 1 mole of $O_2$ weighs 32.00 grams. We can use this to convert the 6.975 moles of $O_2$ to grams:

$6.975 ext{ mol } O_2 imes (32.00 ext{ g } O_2) / (1 ext{ mol } O_2)$

Notice again how the 'mol $O_2$' cancels out, leaving us with grams of $O_2$. Now we just do the multiplication:

$6.975 imes 32.00 = 223.2 ext{ g } O_2$

So, we need 223.2 grams of oxygen gas to completely react with 9.30 moles of aluminum. That's our final answer! We started with moles of aluminum, used stoichiometry to find moles of oxygen, and then converted moles of oxygen to grams. We did it!

Molar Mass: The Bridge Between Moles and Grams

The molar mass of a substance is a fundamental property that serves as a crucial link between the microscopic world of moles and the macroscopic world of grams. It's like having a universal translator that allows us to convert between these two units of measurement. In our case, the molar mass of oxygen gas ($O_2$), which is 32.00 g/mol, tells us that one mole of $O_2$ weighs 32.00 grams. This value is not arbitrary; it's derived from the atomic masses of the elements that make up the molecule, as found on the periodic table. Understanding and utilizing molar mass is essential for quantitative chemistry, as it allows us to accurately measure and predict the amounts of substances involved in chemical reactions.

Setting Up the Final Conversion: Grams, Our Desired Unit

Just as we did with the mole ratio, we set up the final conversion using dimensional analysis to ensure that our units cancel out correctly. We start with the amount of oxygen we have in moles (6.975 mol $O_2$) and multiply it by the molar mass, arranged as a fraction with grams in the numerator and moles in the denominator. This arrangement is deliberate; it allows the 'mol $O_2$' unit to cancel out, leaving us with grams of $O_2$, which is the unit we're seeking. This meticulous attention to units is a hallmark of careful scientific problem-solving. It ensures that our calculations are not only numerically correct but also logically sound.

The Final Calculation: 223.2 Grams of Oxygen Gas

Performing the final calculation, $6.975 ext{ mol } O_2 imes (32.00 ext{ g } O_2) / (1 ext{ mol } O_2) = 223.2 ext{ g } O_2$, brings us to the answer we've been working towards. We've determined that 223.2 grams of oxygen gas are needed to react completely with 9.30 moles of aluminum. This result is not just a number; it's the solution to our problem, the culmination of our step-by-step analysis. It represents a tangible quantity of oxygen gas that would be required in a real-world chemical reaction. This final step underscores the practical application of stoichiometry – its ability to provide quantitative answers to chemical questions.

So, we've successfully navigated the conversion from moles of oxygen to grams of oxygen, using the molar mass as our guide. We've arrived at the final answer, 223.2 grams of $O_2$, and we've demonstrated the power of stoichiometry in solving chemical problems. This journey, from identifying the givens to performing the final calculation, highlights the importance of each step in the process. Understanding the concepts and applying them methodically is the key to success in chemistry!

Conclusion

So, there you have it! We've successfully calculated that 223.2 grams of oxygen gas are needed to completely react with 9.30 moles of aluminum. We started by identifying the givens, used the balanced equation to find the mole ratio, and then converted moles of oxygen to grams using the molar mass. This is a classic stoichiometry problem, and you've now got the tools to tackle similar questions. Remember, the key is to break down the problem into smaller steps, use the balanced equation as your guide, and pay close attention to units. Keep practicing, and you'll become a stoichiometry pro in no time! Keep up the awesome work, guys!