Calculating Lithium Production In The Reaction 2 LiBr + Ba → BaBr₂ + 2 Li
Hey everyone! Let's dive into a cool chemistry problem where we'll figure out how much lithium is produced in a reaction. This is a classic stoichiometry question, and we're going to break it down step by step so it's super clear. So, grab your calculators, and let's get started!
Understanding the Chemical Reaction
In this chemical reaction, we're dealing with the reaction between lithium bromide (LiBr) and barium (Ba) to produce barium bromide (BaBr₂) and lithium (Li). The balanced equation is:
2 LiBr + Ba → BaBr₂ + 2 Li
This balanced chemical equation is super important because it tells us the exact ratio in which the reactants react and the products are formed. In this case, it shows that 2 moles of lithium bromide (LiBr) react with 1 mole of barium (Ba) to produce 1 mole of barium bromide (BaBr₂) and 2 moles of lithium (Li). This mole ratio is the key to solving our problem.
The Significance of Stoichiometry
Stoichiometry, guys, is the bread and butter of quantitative chemistry. It allows us to predict the amounts of reactants and products involved in a chemical reaction. Think of it like a recipe – if you know the ingredients and their proportions, you can figure out how much of the final dish you'll end up with. In chemistry, the balanced equation is our recipe, and the coefficients tell us the proportions in moles. This is crucial for everything from industrial chemical production to lab experiments.
The balanced equation acts as a roadmap, guiding us from the given amount of reactant (barium in this case) to the unknown amount of product (lithium). The coefficients in the balanced equation represent the molar ratios, providing a direct link between the quantities of different substances involved in the reaction. Ignoring the stoichiometry is like baking a cake without measuring ingredients – you might end up with a mess!
Why Balancing Equations Matters
Balancing chemical equations is not just a formality; it's a fundamental principle rooted in the law of conservation of mass. This law states that matter cannot be created or destroyed in a chemical reaction, meaning the number of atoms of each element must be the same on both sides of the equation. A balanced equation ensures that we're accurately representing the conservation of mass, and it provides the correct molar ratios for stoichiometric calculations.
Imagine if the equation were not balanced – we might incorrectly assume that 1 mole of Ba produces 1 mole of Li, when in reality, it produces 2 moles. This error would throw off our entire calculation and lead to a wrong answer. So, always double-check that your equation is balanced before proceeding with any stoichiometric calculations. This small step can save you a lot of trouble and ensure the accuracy of your results.
Problem Breakdown: Grams of Barium to Moles of Lithium
Okay, so we know that 325 grams of barium (Ba) react completely. Our goal is to find out how many moles of lithium (Li) are produced. Here’s the plan:
- Convert grams of barium to moles of barium. To do this, we'll use the molar mass of barium, which you can find on the periodic table. The molar mass of Ba is approximately 137.33 g/mol.
- Use the stoichiometry of the reaction to find moles of lithium. The balanced equation tells us the mole ratio between Ba and Li.
Let’s break down each step with clear calculations and explanations.
Step 1: Converting Grams of Barium to Moles
In this first step, we're tackling a fundamental conversion – transforming grams, a unit of mass, into moles, a unit of amount. The mole is a cornerstone of chemistry, representing a fixed number of particles (6.022 x 10²³ to be exact, also known as Avogadro's number). To make this conversion, we need the molar mass of barium, which is the mass of one mole of Ba atoms. The molar mass is like a conversion factor, bridging the gap between mass and moles. We can find the molar mass of barium on the periodic table; it's approximately 137.33 grams per mole (g/mol). This means that 1 mole of barium has a mass of 137.33 grams.
To convert 325 grams of barium to moles, we'll use the following formula:
Moles = Mass (g) / Molar Mass (g/mol)
Plugging in the values:
Moles of Ba = 325 g / 137.33 g/mol
Calculating this, we get:
Moles of Ba ≈ 2.37 moles
So, 325 grams of barium is approximately equal to 2.37 moles of barium. This conversion is the first critical step in solving our problem because it puts the amount of barium in the language of the balanced equation – moles. From here, we can use the stoichiometry of the reaction to determine how many moles of lithium are produced.
Step 2: Using Stoichiometry to Find Moles of Lithium
Now that we know how many moles of barium we have, we can use the balanced chemical equation to figure out how many moles of lithium are produced. This is where the stoichiometry of the reaction really shines. Remember, the balanced equation tells us the molar ratios between reactants and products. In this case, the equation is:
2 LiBr + Ba → BaBr₂ + 2 Li
This equation tells us that for every 1 mole of barium (Ba) that reacts, 2 moles of lithium (Li) are produced. This 1:2 mole ratio is the key to our calculation. It's like saying that for every one cup of flour you use in a recipe, you get two cookies. The ratio dictates the relationship between the ingredients and the final product.
To find the moles of lithium produced, we’ll use this ratio:
Moles of Li = Moles of Ba × (Moles of Li / Moles of Ba)
We know that we have approximately 2.37 moles of Ba, and the ratio from the balanced equation is 2 moles of Li for every 1 mole of Ba. So, we plug in the values:
Moles of Li = 2.37 moles Ba × (2 moles Li / 1 mole Ba)
Calculating this, we get:
Moles of Li ≈ 4.74 moles
So, approximately 4.74 moles of lithium are produced when 325 grams of barium react completely. This is our final answer, and it highlights the power of stoichiometry in predicting the amounts of products formed in a chemical reaction.
The Final Calculation and Answer
Okay, guys, let's recap! We started with 325 grams of barium (Ba) and wanted to find out how many moles of lithium (Li) were produced in the reaction:
2 LiBr + Ba → BaBr₂ + 2 Li
We broke it down into two main steps:
- Converted grams of barium to moles of barium:
- Moles of Ba = 325 g / 137.33 g/mol ≈ 2.37 moles
- Used the stoichiometry of the reaction to find moles of lithium:
- Moles of Li = 2.37 moles Ba × (2 moles Li / 1 mole Ba) ≈ 4.74 moles
So, the final answer is approximately 4.74 moles of lithium (Li) are produced. Looking at the answer choices, the correct answer is:
C. 4.73 mol
(It's close to our calculated value of 4.74 mol, with a slight difference likely due to rounding along the way.)
Why This Answer Makes Sense
Let's think about why this answer makes sense in the context of the problem. We started with a certain amount of barium and used stoichiometry to predict the amount of lithium produced. The balanced equation showed a 1:2 mole ratio between barium and lithium, meaning we should expect to produce twice as many moles of lithium as we had of barium. Our calculation confirmed this, as we started with approximately 2.37 moles of barium and ended up with approximately 4.74 moles of lithium, which is roughly double. This logical consistency is a good sign that our calculations are on the right track.
Also, consider the molar masses involved. Barium has a molar mass of approximately 137.33 g/mol, while lithium has a much smaller molar mass of approximately 6.94 g/mol. This means that even though we're producing more moles of lithium than we started with of barium, the mass of lithium produced won't be as large as the initial mass of barium. This is another way to check the reasonableness of our answer – it aligns with the relative molar masses and the stoichiometry of the reaction.
Key Takeaways and Tips for Solving Stoichiometry Problems
Awesome job working through this problem, guys! Stoichiometry can seem intimidating at first, but with a clear strategy and a bit of practice, you'll become pros in no time. Here are some key takeaways and tips to keep in mind when tackling stoichiometry problems:
Master the Art of Balancing Equations
Before diving into any calculations, the first and most crucial step is to ensure that your chemical equation is correctly balanced. A balanced equation is the foundation of stoichiometry, providing the accurate mole ratios needed for calculations. An unbalanced equation can lead to incorrect results, so take the time to carefully balance the equation before moving forward. Practice balancing various types of equations, from simple to complex, to build your proficiency in this essential skill.
Convert Grams to Moles – The Golden Rule
The golden rule of stoichiometry is to convert grams to moles as your initial step. Moles are the language of the balanced equation, and stoichiometric calculations are based on molar ratios. Converting grams to moles allows you to work within the framework of the balanced equation and accurately determine the amounts of reactants and products involved. Remember, the formula for converting grams to moles is:
Moles = Mass (g) / Molar Mass (g/mol)
Use Mole Ratios from the Balanced Equation
The coefficients in the balanced equation represent the mole ratios between reactants and products. These mole ratios are your conversion factors for relating the amounts of different substances in the reaction. For example, if the equation shows 2 moles of A reacting with 1 mole of B, the mole ratio of A to B is 2:1. Use these ratios to convert between moles of different substances, ensuring you’re following the stoichiometry of the reaction.
Pay Attention to Units and Significant Figures
In any quantitative problem, paying attention to units and significant figures is essential for accuracy. Make sure your units cancel out correctly during calculations, leaving you with the desired unit for your answer. Use dimensional analysis to track units and prevent errors. Also, be mindful of significant figures throughout the calculation and report your final answer with the appropriate number of significant figures.
Double-Check Your Work and Think Logically
Finally, always double-check your work to ensure you haven't made any calculation errors. Review your steps, units, and significant figures. It's also helpful to think logically about your answer and ask yourself if it makes sense in the context of the problem. Does the magnitude of your answer seem reasonable? Does it align with the stoichiometry of the reaction? Logical thinking can help you catch errors and build confidence in your results.
Practice Makes Perfect: More Stoichiometry Problems
To really nail stoichiometry, practice is key. The more problems you solve, the more comfortable and confident you'll become with the concepts and calculations. So, grab some more stoichiometry problems and put these tips into action. You've got this!
Conclusion
So, there you have it! We successfully calculated the amount of lithium produced in the reaction. Remember, the key is to balance the equation, convert grams to moles, use the mole ratios, and keep track of your units. Keep practicing, and you'll become a stoichiometry superstar! Keep up the awesome work, guys, and happy calculating!
