Calculating Moles Of Oxygen Required To Produce 68.1 G Of Water

Hey there, chemistry enthusiasts! Ever wondered how much oxygen you need to create a specific amount of water in a chemical reaction? Today, we're diving deep into stoichiometry to solve just that! We'll break down the steps, making it super easy to understand. So, grab your lab coats (or just a notepad) and let's get started!

Understanding the Chemical Equation

Before we jump into the calculations, let's first understand the balanced chemical equation we're working with:

$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

This equation tells us the ratio in which propane ($C_3H_8$) reacts with oxygen ($O_2$) to produce carbon dioxide ($CO_2$) and water ($H_2O$). Specifically, it states that 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide and 4 moles of water. This is crucial because it provides the mole ratio needed to solve our problem.

Deciphering the Mole Ratio

The mole ratio is the key to unlocking stoichiometry problems. In our equation, the mole ratio between oxygen ($O_2$) and water ($H_2O$) is 5:4. This means that for every 4 moles of water produced, 5 moles of oxygen are consumed. Think of it as a recipe: if you want to bake 4 cakes, you need 5 cups of flour. Similarly, to produce 4 moles of water, we need 5 moles of oxygen. Understanding this mole ratio is the cornerstone of our calculation.

Why is the Balanced Equation Important?

You might be wondering, why do we even need a balanced equation? Well, the balanced equation ensures that we are adhering to the law of conservation of mass. This fundamental law states that matter cannot be created or destroyed in a chemical reaction. In simpler terms, the number of atoms of each element must be the same on both sides of the equation. Balancing the equation guarantees that we have an accurate representation of the reaction, allowing us to correctly determine the mole ratios and perform stoichiometric calculations.

Real-World Applications

The principles of stoichiometry aren't just confined to textbooks and labs. They have numerous real-world applications. For instance, in industrial chemistry, stoichiometric calculations are essential for optimizing chemical reactions to maximize product yield and minimize waste. In environmental science, they are used to calculate the amount of pollutants produced in combustion processes. Even in cooking, understanding ratios (though not in moles!) is crucial for getting the right flavor and consistency in your dishes. So, mastering stoichiometry is not just about acing exams; it's about understanding the quantitative aspects of the world around us.

Step-by-Step Calculation

Now that we have a solid understanding of the equation and the mole ratio, let's break down the calculation into manageable steps:

Step 1: Convert grams of water to moles

We're given 68.1 g of water ($H_2O$) and we need to convert this to moles. To do this, we'll use the molar mass of water. The molar mass of a substance is the mass of one mole of that substance, usually expressed in grams per mole (g/mol). For water ($H_2O$), the molar mass is approximately 18.015 g/mol.

Molar mass of H2O = (2 × Atomic mass of H) + (1 × Atomic mass of O) Molar mass of H2O = (2 × 1.008 g/mol) + (1 × 16.00 g/mol) Molar mass of H2O = 18.015 g/mol

To convert grams to moles, we use the following formula:

Moles = Mass / Molar mass

Plugging in the values, we get:

Moles of H2O = 68.1 g / 18.015 g/mol Moles of H2O ≈ 3.78 mol

So, 68.1 g of water is approximately equal to 3.78 moles.

Step 2: Use the mole ratio to find moles of oxygen

This is where the balanced equation comes in handy! We know the mole ratio between oxygen ($O_2$) and water ($H_2O$) is 5:4. This means that for every 4 moles of water produced, 5 moles of oxygen are consumed. We can use this ratio to find out how many moles of oxygen are needed to produce 3.78 moles of water.

We can set up a proportion:

(Moles of O2) / (Moles of H2O) = 5 / 4

Let's call the moles of oxygen we're trying to find 'x'. So, the proportion becomes:

x / 3.78 mol = 5 / 4

To solve for x, we multiply both sides of the equation by 3.78 mol:

x = (5 / 4) × 3.78 mol x ≈ 4.73 mol

Therefore, approximately 4.73 moles of oxygen are consumed to produce 3.78 moles (or 68.1 g) of water.

Step 3: Double-Checking Our Work

It's always a good idea to double-check our calculations to ensure accuracy. We can do this by working backward. If 4.73 moles of oxygen are consumed, we should be able to calculate the amount of water produced using the mole ratio. Let's do it!

Using the mole ratio of 4:5 (water to oxygen), we can set up another proportion:

(Moles of H2O) / (Moles of O2) = 4 / 5

Let's call the moles of water produced 'y'. So, the proportion becomes:

y / 4.73 mol = 4 / 5

To solve for y, we multiply both sides of the equation by 4.73 mol:

y = (4 / 5) × 4.73 mol y ≈ 3.78 mol

This matches our initial calculation of 3.78 moles of water, so we can be confident in our answer.

Common Pitfalls to Avoid

Stoichiometry can be tricky, and there are a few common mistakes that students often make. One common error is using an unbalanced equation. Remember, the mole ratios are only accurate if the equation is balanced! Another mistake is incorrectly converting grams to moles or vice versa. Always double-check your molar masses and units. Finally, make sure you're using the correct mole ratio from the balanced equation. Carefully identify the substances you're interested in and use their coefficients to determine the ratio.

Final Answer

So, to produce 68.1 g of water in this reaction, you would need approximately 4.73 moles of oxygen. Great job, guys! You've successfully navigated a stoichiometry problem. Keep practicing, and you'll become a pro in no time!

Wrapping Up and Key Takeaways

We've covered a lot in this article, from understanding the balanced chemical equation to performing the calculations step-by-step. Let's recap the key takeaways:

• Balanced Chemical Equations are Crucial: They provide the accurate mole ratios needed for stoichiometric calculations.
• Mole Ratio is the Key: The mole ratio allows us to convert between moles of different substances in a reaction.
• Grams to Moles Conversion: We use molar mass to convert between grams and moles.
• Practice Makes Perfect: The more you practice stoichiometry problems, the easier they become.

Stoichiometry might seem daunting at first, but with a clear understanding of the concepts and a systematic approach, you can tackle any problem. Keep exploring, keep learning, and keep those chemical reactions balanced!

Practice Problems

Want to put your newfound skills to the test? Here are a couple of practice problems for you to try:

1. How many moles of carbon dioxide are produced when 2.5 moles of propane react completely?
2. If you want to produce 100 g of carbon dioxide, how many moles of oxygen are required?

Try solving these problems using the steps we've discussed. You can share your answers in the comments below, and we can discuss them together.

Happy calculating!