Moles Of Water Produced From 13.35 Moles Of Oxygen

Let's dive into the fascinating world of stoichiometry and explore how we can determine the amount of product formed in a chemical reaction. In this article, we'll break down the process of calculating the moles of water produced from a given amount of oxygen in the reaction $2 H_2 + O_2 ightarrow 2 H_2 O$. This is a classic chemistry problem that highlights the importance of understanding mole ratios and balanced chemical equations. So, grab your thinking caps, and let's get started!

Understanding the Balanced Chemical Equation

The foundation of any stoichiometry problem lies in the balanced chemical equation. This equation provides the crucial mole ratios that we need to solve the problem. In our case, the balanced equation is:

$2 H_2 + O_2 ightarrow 2 H_2 O$

This equation tells us that two moles of hydrogen gas ($H_2$) react with one mole of oxygen gas ($O_2$) to produce two moles of water ($H_2 O$). The coefficients in front of each chemical formula represent the number of moles involved in the reaction. These coefficients are the key to unlocking the mole ratios we'll use for our calculations. It's crucial to have a correctly balanced equation because if the equation is not balanced, the mole ratios will be incorrect, leading to inaccurate results. Balancing chemical equations ensures that the number of atoms of each element is the same on both sides of the equation, adhering to the law of conservation of mass.

Think of the balanced equation as a recipe for a chemical reaction. Just like a recipe specifies the amount of each ingredient needed, a balanced chemical equation specifies the amount of each reactant and product in terms of moles. By understanding the ratios in the equation, we can predict how much product will be formed from a given amount of reactants. This is the essence of stoichiometry, which allows us to make quantitative predictions about chemical reactions. The coefficients in the balanced equation act as conversion factors, allowing us to move between moles of different substances in the reaction.

Determining the Mole Ratio

The mole ratio is the cornerstone of stoichiometric calculations. It allows us to convert between moles of one substance and moles of another in a chemical reaction. From the balanced equation, $2 H_2 + O_2 ightarrow 2 H_2 O$, we can identify several mole ratios. However, for our problem, we are interested in the relationship between oxygen ($O_2$) and water ($H_2 O$).

Looking at the balanced equation, we see that 1 mole of $O_2$ produces 2 moles of $H_2 O$. This gives us the mole ratio:

rac{2 ext{ moles } H_2 O}{1 ext{ mole } O_2}

This ratio is our conversion factor. It tells us that for every 1 mole of oxygen that reacts, 2 moles of water are produced. It's essential to set up the mole ratio correctly, ensuring that the units we want to convert from are in the denominator and the units we want to convert to are in the numerator. In this case, we want to convert from moles of $O_2$ to moles of $H_2 O$, so we place moles of $O_2$ in the denominator and moles of $H_2 O$ in the numerator. The mole ratio is a direct consequence of the law of definite proportions, which states that a chemical compound always contains the same elements in the same proportions by mass. This law underscores the importance of fixed ratios in chemical reactions and is the basis for stoichiometric calculations.

Calculating Moles of Water Produced

Now that we have the mole ratio, we can calculate the moles of water produced from 13.35 moles of oxygen. We'll use the mole ratio as a conversion factor. Start with the given amount of oxygen:

13. 35 moles $O_2$

Multiply this by the mole ratio to convert moles of $O_2$ to moles of $H_2 O$:

13.35 ext{ moles } O_2 imes rac{2 ext{ moles } H_2 O}{1 ext{ mole } O_2}

Notice how the units of moles $O_2$ cancel out, leaving us with moles of $H_2 O$:

$13.35 imes 2 ext{ moles } H_2 O = 26.70 ext{ moles } H_2 O$

Therefore, 26.70 moles of water are produced from 13.35 moles of oxygen. This calculation demonstrates the power of stoichiometry in predicting the amount of product formed in a chemical reaction. By using the mole ratio derived from the balanced chemical equation, we can accurately determine the quantity of water produced. This type of calculation is fundamental in various chemical applications, including industrial processes, research, and even everyday life, such as cooking and baking, where precise ratios of ingredients are essential for successful outcomes.

Practical Implications and Real-World Applications

The calculation we performed isn't just a theoretical exercise; it has significant practical implications. Understanding stoichiometry is crucial in many fields, including:

• Industrial Chemistry: In industrial processes, chemists need to know exactly how much of each reactant to use to produce a desired amount of product. This ensures efficiency and minimizes waste. For example, in the synthesis of ammonia via the Haber-Bosch process, stoichiometric calculations are essential for optimizing the reaction conditions and maximizing ammonia yield.
• Pharmaceuticals: Drug synthesis requires precise control over the amounts of reactants and products. Stoichiometry helps ensure that the correct dosage of a drug is produced, which is critical for patient safety.
• Environmental Science: Stoichiometry is used to calculate the amounts of pollutants produced in chemical reactions, such as combustion. This information is vital for developing strategies to mitigate pollution.
• Research: Chemists use stoichiometry to design experiments and analyze data. It helps them understand reaction mechanisms and predict the outcomes of chemical reactions. For example, in developing new catalysts, stoichiometric calculations help determine the optimal catalyst loading for a given reaction.

Moreover, the water formation reaction itself is vital in various applications. It is a fundamental process in combustion, where the energy released from the reaction is harnessed for power generation. The reaction is also crucial in fuel cells, where hydrogen and oxygen react electrochemically to produce electricity and water. Therefore, a thorough understanding of the stoichiometry of this reaction is essential for advancing these technologies. By mastering stoichiometric principles, chemists and engineers can optimize chemical processes, develop new materials, and solve critical problems in various industries.

Key Takeaways and Summary

Let's recap the key steps we took to solve this problem. First, we started with the balanced chemical equation: $2 H_2 + O_2 ightarrow 2 H_2 O$. This equation provided the crucial mole ratio between oxygen and water. Next, we identified the mole ratio of 2 moles $H_2 O$ to 1 mole $O_2$. This ratio served as our conversion factor. Finally, we multiplied the given amount of oxygen (13.35 moles) by the mole ratio to calculate the moles of water produced, which was 26.70 moles.

This problem highlights the importance of understanding stoichiometry and using balanced chemical equations to solve quantitative problems in chemistry. Stoichiometry is not just about numbers; it's about understanding the relationships between substances in a chemical reaction and predicting the outcomes of these reactions. By mastering these concepts, you can tackle a wide range of chemical calculations and gain a deeper appreciation for the quantitative nature of chemistry. The ability to perform stoichiometric calculations is a fundamental skill for anyone studying chemistry or working in a related field.

In conclusion, we've successfully calculated the moles of water produced from a given amount of oxygen using stoichiometry. This process involved understanding the balanced chemical equation, determining the mole ratio, and applying it as a conversion factor. Remember, practice makes perfect, so keep working on stoichiometry problems to solidify your understanding. For further learning and to deepen your understanding of stoichiometry, consider exploring resources like Khan Academy's Chemistry section, which offers comprehensive lessons and practice exercises on this essential topic.