Moles Of Oxygen Needed To React With Hydrogen

Understanding stoichiometry is crucial in chemistry, especially when dealing with chemical reactions. In this article, we will explore how to determine the number of moles of oxygen required to completely react with a given amount of hydrogen, using a balanced chemical equation as our guide. Let's dive into the details and solve this fascinating problem together.

Understanding the Balanced Equation

The foundation of our calculation lies in the balanced chemical equation:

$2 H_2 + O_2 \rightarrow 2 H_2 O$

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 ($H_2O$). The coefficients in front of each chemical formula represent the stoichiometric ratios, which are vital for determining the molar relationships between reactants and products.

To truly grasp the significance of this equation, let’s break it down further. The equation is not just a symbolic representation; it's a quantitative statement about the reaction. It explicitly states that for every two molecules of hydrogen that react, one molecule of oxygen is consumed. This 2:1 ratio is the key to solving our problem. In the macroscopic world, we deal with moles, which are simply a scaled-up version of this molecular relationship. One mole of a substance contains Avogadro's number ($6.022 imes 10^{23}$) of particles, so the mole ratios in the balanced equation hold true at this scale as well. This understanding of the balanced equation sets the stage for us to calculate the exact amount of oxygen needed to react with a specific quantity of hydrogen.

The coefficients in the balanced equation are not arbitrary numbers; they are the result of ensuring that the number of atoms of each element is conserved during the chemical reaction. In other words, matter is neither created nor destroyed, only rearranged. For instance, in the given equation, we have four hydrogen atoms on the left (2 molecules of $H_2$, each with 2 H atoms) and four hydrogen atoms on the right (2 molecules of $H_2O$, each with 2 H atoms). Similarly, there are two oxygen atoms on the left (1 molecule of $O_2$) and two oxygen atoms on the right (2 molecules of $H_2O$, each with 1 O atom). This conservation of atoms is a fundamental principle in chemistry and a cornerstone of stoichiometry.

Problem Setup: Moles of Hydrogen and Oxygen

We are given that we have 27.4 moles of $H_2$. Our goal is to find out how many moles of $O_2$ are required to react completely with this amount of hydrogen. The balanced equation provides the crucial link between these two quantities.

The problem at hand is a classic stoichiometry question, which hinges on the molar ratios derived from the balanced chemical equation. We are starting with a known quantity of one reactant, hydrogen, and we need to determine the quantity of another reactant, oxygen, required for complete reaction. The phrase "completely react" is a significant indicator that we are dealing with a stoichiometric problem, where the reactants will be consumed in the exact proportions specified by the balanced equation. This means there will be no leftover hydrogen or oxygen once the reaction is complete. Setting up the problem correctly is half the battle in solving it. We need to clearly identify what we know (27.4 moles of $H_2$) and what we want to find (moles of $O_2$). This structured approach will guide us in applying the stoichiometric principles effectively.

Before we proceed with the calculation, it's helpful to visualize the reaction conceptually. Imagine we have a container filled with 27.4 moles of hydrogen gas. To ensure all of this hydrogen reacts to form water, we need to add the precise amount of oxygen. Too little oxygen, and some hydrogen will be left unreacted. Too much oxygen, and we'll have excess oxygen after all the hydrogen is consumed. The balanced equation allows us to calculate this "Goldilocks" amount of oxygen – not too much, not too little, but just right. This conceptual understanding helps reinforce the importance of stoichiometry in ensuring reactions proceed efficiently and predictably.

Applying the Stoichiometric Ratio

From the balanced equation, we know that 2 moles of $H_2$ react with 1 mole of $O_2$. This gives us the stoichiometric ratio:

$\frac{1 ext{ mol } O_2}{2 ext{ mol } H_2}$

This ratio is the conversion factor we will use to convert moles of $H_2$ to moles of $O_2$. To do this, we multiply the given moles of $H_2$ by this ratio:

$27.4 ext{ mol } H_2 imes \frac{1 ext{ mol } O_2}{2 ext{ mol } H_2}$

The beauty of using stoichiometric ratios lies in their ability to act as conversion factors, allowing us to move seamlessly between quantities of different substances in a chemical reaction. In this case, we're using the ratio to convert from the moles of hydrogen we have to the moles of oxygen we need. The units play a crucial role here; by setting up the calculation with moles of $H_2$ in the numerator and denominator, they cancel out, leaving us with moles of $O_2$, which is exactly what we're looking for. This dimensional analysis is a powerful tool in chemistry for ensuring we're performing the correct calculations and arriving at the right units.

To truly appreciate the power of this stoichiometric ratio, consider the alternative: trying to solve this problem without it. We might intuitively understand that we need less oxygen than hydrogen, but without the precise ratio, we'd be left guessing. The balanced equation and the stoichiometric ratio it provides give us a rigorous, quantitative method for determining the exact amount of reactants needed. This precision is essential in many chemical applications, from industrial processes to laboratory experiments, where precise control over reaction conditions is critical.

Calculation and Solution

Performing the calculation:

$27.4 ext{ mol } H_2 imes \frac{1 ext{ mol } O_2}{2 ext{ mol } H_2} = 13.7 ext{ mol } O_2$

Therefore, 13.7 moles of $O_2$ are required to completely react with 27.4 moles of $H_2$.

The arithmetic in this calculation is straightforward, but the underlying concept is profound. We've essentially used the balanced chemical equation as a recipe, scaling the ingredients (reactants) according to the proportions specified. In this case, for every two units of hydrogen, we need one unit of oxygen. Since we have 27.4 moles of hydrogen, we logically need half that amount of oxygen, which is 13.7 moles. This simple calculation demonstrates the power of stoichiometry in quantifying chemical reactions.

To further solidify our understanding, let's consider what the answer means in a practical context. If we were conducting this reaction in a laboratory, we would need to carefully measure out 27.4 moles of hydrogen gas and 13.7 moles of oxygen gas to ensure the reaction proceeds efficiently and completely. Using significantly more or less oxygen could lead to incomplete reaction, wasted reactants, or even the formation of unwanted byproducts. This highlights the importance of stoichiometry in chemical synthesis and experimentation.

Choosing the Correct Answer

Looking at the options provided:

A. 6.8 mol B. 13.7 mol C. 54.8 mol D. 109.6 mol

The correct answer is B. 13.7 mol.

It's crucial not only to arrive at the correct numerical answer but also to understand why the other options are incorrect. Option A, 6.8 mol, represents half the correct amount, suggesting a misunderstanding of the 2:1 ratio or perhaps dividing by 4 instead of 2. Option C, 54.8 mol, is twice the amount of hydrogen, indicating a possible mix-up in the stoichiometric coefficients or a failure to correctly apply the ratio. Option D, 109.6 mol, is four times the correct answer, which could arise from multiplying instead of dividing by 2, or a combination of errors. By analyzing these incorrect options, we can pinpoint common mistakes and reinforce the correct approach to solving stoichiometry problems.

Choosing the correct answer is the final step in the problem-solving process, but it's not just about selecting the right letter. It's about confirming that the answer makes sense in the context of the problem and that we've followed a logical and accurate process to arrive at it. Before settling on an answer, it's always a good practice to double-check our calculations, revisit the balanced equation, and ensure that we've addressed the question that was asked. This thoroughness will not only lead to correct answers but also deepen our understanding of the underlying chemical principles.

Conclusion

In conclusion, to completely react with 27.4 moles of $H_2$, 13.7 moles of $O_2$ are required, based on the balanced equation $2 H_2 + O_2 \rightarrow 2 H_2 O$. Understanding and applying stoichiometric ratios is essential for solving these types of problems. Remember, the balanced equation is your roadmap to accurate calculations in chemistry!

Mastering stoichiometry is a fundamental skill in chemistry, and it's one that opens the door to understanding more complex chemical concepts. The ability to quantitatively relate reactants and products in a chemical reaction is crucial for a wide range of applications, from designing chemical syntheses to analyzing reaction yields. By working through problems like this one, we develop not only our problem-solving skills but also our conceptual understanding of chemistry. The balanced equation is more than just a symbolic representation; it's a powerful tool that allows us to predict and control chemical reactions.

To further enhance your understanding of stoichiometry, consider exploring additional resources and practice problems. Many online platforms and textbooks offer a wealth of information and exercises on this topic. The more you practice, the more comfortable and confident you'll become in applying these principles. Remember, chemistry is a subject that builds upon itself, so a strong foundation in stoichiometry will serve you well in your future studies. Keep practicing, keep exploring, and keep the balanced equation in mind!

For further learning on stoichiometry, you can visit Khan Academy's Stoichiometry Section. This is a trusted website that offers comprehensive lessons and practice exercises.