Calculating Oxygen Needed For A Chemical Reaction

Hey everyone! Let's dive into a cool chemistry problem. We've got a reaction where $C_6H_{12}O_6(s)$ (that's glucose, guys!) reacts with $O_2(g)$ (oxygen gas) to produce $CO_2(g)$ (carbon dioxide) and $H_2O(l)$ (water). The really interesting bit? We know that a whopping $9.81 \times 10^{25}$ molecules of $CO_2$ are produced. Our mission, should we choose to accept it, is to figure out how much $O_2$ (in kilograms) is needed for this reaction. This isn't just about plugging numbers into a formula; it's about understanding the heart of chemical reactions, the mole concept, and the power of stoichiometry. So, buckle up; we're about to explore the fascinating world of chemical calculations! This is where we learn how to use the balanced chemical equation, Avogadro's number, and molar masses to solve real-world problems. By the end, you'll be able to confidently tackle similar stoichiometry problems.

Understanding the Chemical Reaction

First things first, we need to make sure we've got the balanced chemical equation. The unbalanced equation is: $C_6H_{12}O_6(s) + O_2(g) \longrightarrow CO_2(g) + H_2O(l)$. This reaction, in its simplest form, represents cellular respiration—how our bodies (and many others!) get energy from glucose. But, like any good chef, we need to balance the ingredients to make sure everything's just right. Balancing the equation is super important because it tells us the mole ratio of reactants and products. Let's balance it: $C_6H_{12}O_6(s) + 6O_2(g) \longrightarrow 6CO_2(g) + 6H_2O(l)$. See how that works? Now we know that for every one mole of glucose, we need six moles of oxygen to produce six moles of carbon dioxide and six moles of water. This balanced equation is our roadmap. It tells us the ratio in which the reactants combine and the products form. Knowing this ratio is absolutely crucial for all our calculations. We'll be using this balanced equation throughout our journey. Understanding the coefficients (the big numbers in front of the molecules) is key. They tell us how many moles of each substance are involved in the reaction. For example, the coefficient of $O_2$ is 6, indicating that 6 moles of oxygen are needed to react with 1 mole of glucose.

Converting Molecules to Moles: The First Step

Alright, we've got the balanced equation, and now we know the mole ratio. The next step in our adventure is converting the number of $CO_2$ molecules we have ($9.81 \times 10^{25}$) into moles. This is where Avogadro's number comes into play. Avogadro's number tells us that one mole of any substance contains $6.022 \times 10^{23}$ particles (molecules, atoms, etc.). It's like saying a dozen always has 12 items. So, to convert from molecules to moles, we divide the number of molecules by Avogadro's number. Let's do it: Moles of $CO_2 = \frac{9.81 \times 10^{25} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 162.89 \text{ mol}$. We have approximately 162.89 moles of $CO_2$ produced in this reaction. This conversion is a fundamental step in stoichiometry. It allows us to move from the microscopic world (molecules) to the macroscopic world (moles), where we can apply the balanced chemical equation to calculate the quantities of reactants needed or products formed. Remember, understanding how to convert between molecules and moles is vital for solving any stoichiometry problem. You'll be using Avogadro's number a lot, so get comfy with it!

Using the Mole Ratio: Linking $CO_2$ and $O_2$

Now we know how many moles of $CO_2$ are produced. The next exciting part is using the mole ratio from our balanced equation to figure out how many moles of $O_2$ were used. Look back at the balanced equation: $C_6H_{12}O_6(s) + 6O_2(g) \longrightarrow 6CO_2(g) + 6H_2O(l)$. The mole ratio between $CO_2$ and $O_2$ is 6:6, which simplifies to 1:1. This means that for every 1 mole of $CO_2$ produced, we need 1 mole of $O_2$. So, if we have 162.89 moles of $CO_2$, we also need 162.89 moles of $O_2$. Easy peasy! The mole ratio is the heart and soul of stoichiometry. It allows us to convert between the amounts of different substances in a chemical reaction. Without the mole ratio, we wouldn't be able to relate the amount of carbon dioxide produced to the amount of oxygen consumed. Always pay close attention to the balanced equation and the coefficients to determine the correct mole ratio. This step highlights the importance of the balanced equation; without it, we wouldn't be able to establish this crucial link between the reactants and products.

From Moles of $O_2$ to Grams: Finding the Mass

We're getting closer! Now we know we have 162.89 moles of $O_2$. But the question asks for the mass in kilograms. So, we need to convert moles of $O_2$ to grams, and then grams to kilograms. To do this, we'll use the molar mass of $O_2$. The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). To find the molar mass of $O_2$, we look at the periodic table. Oxygen (O) has a molar mass of approximately 16.00 g/mol. Since $O_2$ has two oxygen atoms, its molar mass is $2 \times 16.00 = 32.00$ g/mol. Now, we convert moles of $O_2$ to grams: Grams of $O_2 = 162.89 \text{ mol} \times 32.00 \text{ g/mol} \approx 5212.48 \text{ g}$.

This conversion is all about using the molar mass to bridge the gap between moles and grams. Molar mass is a fundamental concept in chemistry, allowing us to convert between mass and the amount of substance (in moles). It's super important to understand how to use the periodic table to find the molar masses of different elements and compounds. Remember to always use the correct units (g/mol) to ensure your calculations are accurate. Pay close attention to the number of atoms of each element in the molecule when calculating molar mass.

Converting Grams to Kilograms: The Final Stretch

We're in the home stretch, guys! We have the mass of $O_2$ in grams (5212.48 g), and we need to convert it to kilograms. This is a simple unit conversion. We know that 1 kilogram (kg) equals 1000 grams (g). So, to convert grams to kilograms, we divide by 1000: Kilograms of $O_2 = \frac{5212.48 \text{ g}}{1000 \text{ g/kg}} \approx 5.21 \text{ kg}$. Boom! We did it! We've successfully calculated that approximately 5.21 kg of $O_2$ is required for this reaction. Always double-check your units at the end of a calculation to make sure they're what you need. This final step is a crucial reminder that we often need to convert between different units to get our final answer in the desired format. Unit conversions are a fundamental skill in science and engineering. Make sure you understand how to move between different units of measurement, such as grams and kilograms, using appropriate conversion factors. Always be mindful of significant figures throughout your calculations, and round your final answer to the correct number of significant figures.

Conclusion: You've Got This!

We've successfully solved our chemistry problem, starting from the number of $CO_2$ molecules produced and working our way to the mass of $O_2$ required. We used the balanced chemical equation, Avogadro's number, molar masses, and the mole ratio to do it. Stoichiometry can seem tricky at first, but with practice, you'll become a pro at these types of calculations. Remember the key steps: balance the equation, convert to moles, use the mole ratio, and convert to the desired units. Keep practicing, and you'll be acing chemistry problems in no time! Keep in mind that understanding each step is vital to mastering stoichiometry. The ability to break down the problem into smaller, manageable parts makes the overall calculation much easier to tackle. If you can understand the principles discussed in this article, you are well on your way to mastering stoichiometry. Congratulations on completing this challenging problem! You've learned how to apply key chemical concepts to solve a real-world problem. Keep up the amazing work! If you have any further questions about these calculations, feel free to ask. There are tons of problems to practice with, so you will improve and be able to solve them independently.