Moles Of Oxygen Needed For 68.1g Water Production

Aug 5, 2025 by Pedro Alvarez 50 views

How Many Moles of Oxygen are Needed to Produce 68.1 g of Water?

Hey there, chemistry enthusiasts! Let's dive into a stoichiometry problem that'll help us understand the relationship between reactants and products in a chemical reaction. Today, we're tackling the question: How many moles of oxygen ( $O_2$ ) would be consumed to produce 68.1 g of water ( $H_2O$ ) in the following reaction?

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

This equation represents the combustion of propane ( $C_3H_8$ ), a common fuel, with oxygen to produce carbon dioxide ( $CO_2$ ) and water. To solve this problem, we'll need to use stoichiometry, the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. Let's break it down step-by-step.

1. Understanding Stoichiometry: The Foundation of Our Calculation

Before we jump into the calculations, let's make sure we're all on the same page about stoichiometry. Stoichiometry is like the recipe book of chemistry. It allows us to predict how much of a substance we need or will produce in a chemical reaction. The balanced chemical equation is the recipe itself, providing the mole ratios between reactants and products. These mole ratios are the key to converting between amounts of different substances in a reaction.

In our equation, $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$ , the coefficients in front of each chemical formula tell us the number of moles involved in the reaction. For instance, the '5' in front of $O_2$ means that 5 moles of oxygen are required for the reaction to occur as written. Similarly, the '4' in front of $H_2O$ tells us that 4 moles of water are produced.

Why is this important? Because we're trying to find out how many moles of oxygen are needed to produce a specific amount of water (68.1 g). We can't directly convert grams of water to moles of oxygen. We need to use the mole ratio from the balanced equation as a bridge.

Converting Grams to Moles: The First Step

Grams are a common way to measure substances in the lab, but stoichiometry works with moles. So, our first task is to convert the given mass of water (68.1 g) into moles. To do this, we'll need the molar mass of water.

The molar mass of a compound is the mass of one mole of that compound, usually expressed in grams per mole (g/mol). We can calculate the molar mass by adding up the atomic masses of all the atoms in the molecule. For water ( $H_2O$ ), we have:

2 hydrogen atoms (H), each with an atomic mass of approximately 1.01 g/mol
1 oxygen atom (O), with an atomic mass of approximately 16.00 g/mol

So, the molar mass of water is (2 * 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol.

Now we can use this to convert grams of water to moles:

Moles of $H_2O$ = (Mass of $H_2O$ ) / (Molar mass of $H_2O$ ) Moles of $H_2O$ = 68.1 g / 18.02 g/mol Moles of $H_2O$ ≈ 3.78 mol

So, we have approximately 3.78 moles of water.

2. Using the Mole Ratio: Bridging Water and Oxygen

Now that we know how many moles of water are produced, we can use the mole ratio from the balanced equation to find out how many moles of oxygen were consumed. The balanced equation tells us that 5 moles of $O_2$ produce 4 moles of $H_2O$ . This gives us the following mole ratio:

(5 moles $O_2$ ) / (4 moles $H_2O$ )

We can use this ratio as a conversion factor. We want to find moles of $O_2$ , and we know moles of $H_2O$ , so we'll multiply the moles of $H_2O$ by the mole ratio:

Moles of $O_2$ = (Moles of $H_2O$ ) * (Mole ratio) Moles of $O_2$ = 3.78 mol $H_2O$ * (5 moles $O_2$ / 4 moles $H_2O$ )

Notice how the units "moles $H_2O$ " cancel out, leaving us with moles of $O_2$ . This is a crucial part of dimensional analysis, ensuring we're doing the calculation correctly.

3. Calculating the Moles of Oxygen: The Final Step

Now, let's do the math:

Moles of $O_2$ = 3.78 mol * (5/4) Moles of $O_2$ = 3.78 mol * 1.25 Moles of $O_2$ ≈ 4.73 mol

Therefore, approximately 4.73 moles of oxygen would be consumed to produce 68.1 g of water in this reaction. Guys, that's our final answer!

Common Mistakes to Avoid in Stoichiometry Problems

Stoichiometry can be tricky, and there are a few common pitfalls students often encounter. Let's go over some of them to help you avoid these errors:

Not Balancing the Equation: This is the most crucial step! If your equation isn't balanced, the mole ratios will be incorrect, and your answer will be wrong. Always double-check that the number of atoms of each element is the same on both sides of the equation.
Using the Wrong Mole Ratio: Make sure you're using the correct coefficients from the balanced equation to create the mole ratio. Double-check which substances you're converting between and use the corresponding coefficients.
Mixing Up Molar Mass and Mole Ratio: Molar mass is used to convert between grams and moles of the same substance, while the mole ratio is used to convert between moles of different substances in a reaction. Don't mix them up!
Incorrect Units: Always include units in your calculations and make sure they cancel out correctly. This will help you identify errors and ensure you're on the right track.
Rounding Errors: Avoid rounding intermediate calculations too much. It's best to keep extra digits during calculations and only round the final answer to the appropriate number of significant figures.

4. Real-World Applications of Stoichiometry: More Than Just Calculations

Stoichiometry isn't just a theoretical concept confined to textbooks and classrooms. It's a fundamental tool used in various real-world applications, impacting industries from pharmaceuticals to manufacturing. Here are a few examples:

Pharmaceutical Industry: Stoichiometry plays a crucial role in drug synthesis. Pharmacists and chemists use stoichiometric calculations to determine the precise amounts of reactants needed to produce a specific quantity of a drug. This ensures the efficiency of the synthesis and minimizes waste.
Chemical Manufacturing: In chemical plants, stoichiometry is essential for scaling up chemical reactions from the laboratory to industrial production. Engineers use stoichiometric principles to optimize reaction conditions, maximize product yield, and control costs.
Environmental Science: Stoichiometry is used in environmental studies to understand and manage pollution. For example, it can be used to calculate the amount of a neutralizing agent needed to treat acidic wastewater or to determine the amount of greenhouse gases produced by a combustion process.
Food Industry: Stoichiometry is applied in food processing to ensure the correct proportions of ingredients are used in recipes. This is crucial for maintaining the taste, texture, and nutritional value of food products.
Combustion Analysis: Stoichiometry is fundamental in combustion analysis, where the elemental composition of a substance is determined by burning it and measuring the amounts of the products formed. This technique is used in various fields, including material science and forensic science.

By understanding stoichiometry, scientists and engineers can design and control chemical processes more effectively, leading to more efficient and sustainable practices.

5. Practice Problems: Sharpen Your Stoichiometry Skills

To truly master stoichiometry, practice is key. Here are a couple of practice problems you can try to test your understanding:

Problem 1:

Consider the following reaction:

$2H_2 + O_2 \rightarrow 2H_2O$

If you have 10 grams of hydrogen ( $H_2$ ), how many grams of oxygen ( $O_2$ ) are needed for complete reaction?

Problem 2:

Ammonia ( $NH_3$ ) is produced by the reaction of nitrogen ( $N_2$ ) and hydrogen ( $H_2$ ):

$N_2 + 3H_2 \rightarrow 2NH_3$

If you want to produce 34 grams of ammonia, how many moles of nitrogen gas do you need?

Tips for Solving:

Make sure the equation is balanced.
Convert grams to moles using molar mass.
Use the mole ratio from the balanced equation to convert between substances.
Convert back to grams if needed.

Solving these problems will help you solidify your understanding of stoichiometry and build confidence in tackling more complex chemical calculations.

Conclusion: Stoichiometry Unlocked!

So, guys, we've successfully calculated that approximately 4.73 moles of oxygen are needed to produce 68.1 g of water in the given reaction. We've also explored the fundamental concepts of stoichiometry, common mistakes to avoid, and real-world applications of this crucial chemical principle. Stoichiometry is a powerful tool for understanding and predicting chemical reactions. By mastering these concepts, you'll be well-equipped to tackle a wide range of chemistry problems and appreciate the quantitative nature of the world around us. Keep practicing, keep exploring, and keep that chemistry curiosity burning! Remember, every chemical reaction tells a story, and stoichiometry helps us decipher it.