Calculate Moles Of Oxygen From Potassium Chlorate

Hey everyone! Today, we're diving into a classic stoichiometry problem: figuring out how many moles of oxygen ($O_2$) are produced when potassium chlorate ($KClO_3$) decomposes. This is a common type of question in chemistry, and understanding how to solve it is super important for mastering chemical reactions and calculations. So, let's break it down step by step, making sure everything is crystal clear.

Understanding the Balanced Chemical Equation

First things first, we need to take a good look at the balanced chemical equation provided:

$2 KClO_3 ightarrow 2 KCl + 3 O_2$

This equation is the heart and soul of our calculation. It tells us exactly how many moles of each substance are involved in the reaction. In plain English, it's saying that when 2 moles of potassium chlorate ($KClO_3$) decompose, they produce 2 moles of potassium chloride ($KCl$) and, crucially for us, 3 moles of oxygen gas ($O_2$). Think of it like a recipe: if you double the ingredients, you double the output. The coefficients (the numbers in front of the chemical formulas) are super important because they give us the mole ratios we need to solve the problem. So, understanding this balanced equation is absolutely fundamental. Without it, we'd be flying blind. This equation acts as a roadmap, guiding us from the amount of reactant we start with ($KClO_3$) to the amount of product we're interested in ($O_2$). It’s like having a secret code that unlocks the relationship between the chemicals in the reaction. Make sure you always start by double-checking the equation is balanced because even a small mistake here can throw off your entire calculation. We need to account for every atom present to ensure mass is conserved, and that's precisely what a balanced equation achieves. It ensures that what goes in (reactants) equals what comes out (products) in terms of atoms and mass, providing the foundation for accurate mole calculations.

Setting up the Mole Ratio

Now that we've got the balanced equation down, let's zoom in on the key players for our problem: potassium chlorate ($KClO_3$) and oxygen ($O_2$). The equation tells us that 2 moles of $KClO_3$ produce 3 moles of $O_2$. This is our mole ratio, and it's the golden ticket to solving this problem. We can write this ratio as a fraction: (3 moles $O_2$) / (2 moles $KClO_3$). This fraction acts as a conversion factor, allowing us to switch from moles of $KClO_3$ to moles of $O_2$. Imagine it as a bridge, smoothly connecting the amount of reactant we have to the amount of product we can make. Understanding this ratio is crucial because it dictates the quantitative relationship between the substances involved in the reaction. If we were baking a cake, this ratio would be like knowing how many eggs we need for every cup of flour. Getting the ratio right is paramount to getting the correct answer. This mole ratio isn't just some abstract number; it’s a direct reflection of the stoichiometry of the reaction, telling us precisely how the amounts of different substances are related. So, whenever you're tackling a stoichiometry problem, always start by identifying the relevant mole ratio from the balanced chemical equation. It's the cornerstone of your calculation.

Performing the Calculation

Okay, guys, we're in the home stretch! We know we're starting with 0.046 moles of $KClO_3$, and we've got our mole ratio of (3 moles $O_2$) / (2 moles $KClO_3$). Now, it's just a matter of putting them together. To find the moles of $O_2$ produced, we'll multiply the moles of $KClO_3$ by our mole ratio:

Moles of $O_2$ = 0.046 moles $KClO_3$ * (3 moles $O_2$) / (2 moles $KClO_3$)

Notice how the units of "moles $KClO_3$" cancel out, leaving us with moles of $O_2$, which is exactly what we want! This unit cancellation is a super handy way to double-check that we've set up our calculation correctly. If your units don't cancel out to give you the units you're looking for, that's a big red flag! Now, let's crunch the numbers:

Moles of $O_2$ = (0.046 * 3) / 2 = 0.069 moles

So, we've found that 0.046 moles of $KClO_3$ will produce 0.069 moles of $O_2$. That wasn't so bad, was it? The key here is to take it one step at a time, making sure you understand each step before moving on. It's like building with LEGO bricks; each brick (or step) needs to be firmly in place before you add the next one. And remember, always double-check your work to avoid silly mistakes. A small error in calculation can lead to a wrong final answer, so accuracy is crucial. Stoichiometry problems can seem intimidating at first, but with practice and a methodical approach, you'll be solving them like a pro in no time!

The Answer

Therefore, the correct answer is A. 0.069 mol. We've successfully navigated this stoichiometry problem, and hopefully, you've gained a clearer understanding of how to tackle similar questions. Remember, chemistry is all about understanding the relationships between substances, and mole ratios are a powerful tool in that endeavor. Keep practicing, keep asking questions, and you'll become a chemistry whiz in no time! Don't be discouraged if you don't get it right away; practice makes perfect, and every problem you solve is a step forward. The journey to mastering chemistry is a marathon, not a sprint, so be patient with yourself and celebrate your progress along the way.

Key Takeaways for Stoichiometry Success

Before we wrap up, let's recap the essential steps for solving stoichiometry problems like this one. These key takeaways will serve you well as you tackle more complex chemical calculations in the future.

1. Start with a Balanced Chemical Equation: This is the bedrock of all stoichiometric calculations. A balanced equation provides the mole ratios that link reactants and products, so always make this your first step.
2. Identify the Mole Ratio: Once you have a balanced equation, pinpoint the mole ratio between the substances you're interested in. This ratio acts as your conversion factor, allowing you to move between moles of different chemicals.
3. Set Up the Calculation Correctly: Multiply the given amount (in moles) of one substance by the mole ratio, ensuring that the units cancel out to give you the desired units. This step is where careful attention to detail is crucial.
4. Crunch the Numbers: Perform the calculation accurately, paying attention to significant figures and rounding rules.
5. Double-Check Your Work: Always review your setup and calculations to catch any errors. Unit cancellation is a great way to verify that your approach is sound.
6. Practice, Practice, Practice: The more you work through stoichiometry problems, the more comfortable and confident you'll become. Seek out examples, try different variations, and don't be afraid to ask for help.

By mastering these steps, you'll be well-equipped to handle a wide range of stoichiometry challenges. So, go forth and conquer those chemical calculations!

Moles of Oxygen from Potassium Chlorate Calculation - A Chemistry Guide