Calculating Moles Of Barium Hydroxide Required To React With Hydrogen Bromide

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Have you ever wondered how much of one chemical you need to react completely with another? In chemistry, this is a common question, and it involves understanding stoichiometry. Stoichiometry, guys, is just a fancy word for the math behind chemical reactions. It's all about the relationships between the amounts of reactants and products in a chemical equation. In this article, we're going to dive into a specific example: figuring out how many moles of barium hydroxide, Ba(OH)₂, are needed to react with 117 grams of hydrogen bromide, HBr. This is a classic stoichiometry problem, and by the end of this, you’ll be able to tackle similar calculations with confidence.

The reaction we're focusing on is:

2 HBr + Ba(OH)₂ → BaBr₂ + 2 H₂O

This balanced chemical equation tells us a lot. It's like a recipe for a chemical reaction. It says that two moles of hydrogen bromide (HBr) react with one mole of barium hydroxide (Ba(OH)₂) to produce one mole of barium bromide (BaBr₂) and two moles of water (H₂O). The numbers in front of the chemical formulas (like the '2' in front of HBr) are called stoichiometric coefficients, and they are super important for our calculations. These coefficients are the key to unlocking the mole ratios, which will guide us to the final answer. Without a balanced equation, we'd be stumbling in the dark, so always make sure your equation is balanced before you start any stoichiometry problem!

Now, let's break down how to solve this problem step-by-step. We'll start with what we know: we have 117 grams of HBr. But, to use the balanced equation, we need to convert grams to moles. Moles, guys, are the chemists' favorite way to measure amounts of substances because they directly relate to the number of molecules or atoms. To convert grams to moles, we need the molar mass of HBr. The molar mass is the mass of one mole of a substance, and it's usually expressed in grams per mole (g/mol). You can find the molar mass by adding up the atomic masses of each element in the compound from the periodic table. For HBr, it's the atomic mass of hydrogen (approximately 1.01 g/mol) plus the atomic mass of bromine (approximately 79.90 g/mol), which gives us a molar mass of about 80.91 g/mol for HBr. This molar mass is our conversion factor, the bridge that takes us from the world of grams to the world of moles.

Step-by-Step Calculation

Let's dive into the step-by-step calculation to determine how many moles of barium hydroxide are required to react with 117 g of hydrogen bromide.

Step 1: Convert grams of HBr to moles of HBr

As we discussed earlier, the first crucial step is to convert the given mass of hydrogen bromide (HBr) into moles. This conversion is essential because the balanced chemical equation relates the amounts of substances in terms of moles, not grams. To perform this conversion, we'll use the molar mass of HBr, which, as we calculated, is approximately 80.91 g/mol. Remember, the molar mass acts as a bridge, connecting the mass of a substance to the number of moles. It tells us the mass of one mole of that substance.

The formula for converting grams to moles is:

Moles = Grams / Molar mass

In our case, we have 117 grams of HBr, and the molar mass of HBr is 80.91 g/mol. Plugging these values into the formula, we get:

Moles of HBr = 117 g / 80.91 g/mol ≈ 1.446 moles

So, we have approximately 1.446 moles of HBr. This is a key intermediate result. We've now expressed the amount of HBr in the units that our balanced equation understands: moles. This conversion is like translating a recipe from ounces to cups; we're putting the measurement in the right terms for the next step.

Step 2: Use the Stoichiometric Ratio to Find Moles of Ba(OH)₂

Now that we know the number of moles of HBr, we can use the balanced chemical equation to determine the number of moles of barium hydroxide (Ba(OH)₂) needed to react completely. This is where the coefficients in the balanced equation come into play. They provide the stoichiometric ratio, which is the mole ratio between the reactants and products.

Looking at the balanced equation:

2 HBr + Ba(OH)₂ → BaBr₂ + 2 H₂O

We see that 2 moles of HBr react with 1 mole of Ba(OH)₂. This gives us the stoichiometric ratio we need. It's like saying, "For every two slices of bread, you need one slice of cheese." The ratio is the key to scaling the recipe up or down.

We can express this ratio as a fraction: (1 mole Ba(OH)₂) / (2 moles HBr). This fraction acts as a conversion factor, allowing us to convert from moles of HBr to moles of Ba(OH)₂. It's important to set up the fraction correctly, ensuring that the units we want to cancel out (moles of HBr) are in the denominator, and the units we want to find (moles of Ba(OH)₂) are in the numerator.

To find the moles of Ba(OH)₂ required, we multiply the moles of HBr we calculated in step 1 by this stoichiometric ratio:

Moles of Ba(OH)₂ = 1.446 moles HBr * (1 mole Ba(OH)₂ / 2 moles HBr)

The "moles HBr" units cancel out, leaving us with:

Moles of Ba(OH)₂ ≈ 0.723 moles

So, approximately 0.723 moles of barium hydroxide (Ba(OH)₂) are required to react completely with 1.446 moles (or 117 grams) of hydrogen bromide (HBr). This is our final answer! We've successfully navigated the stoichiometric landscape, using the balanced equation as our map and the mole ratios as our compass.

Final Answer

Therefore, approximately 0.723 moles of barium hydroxide, Ba(OH)₂, would be required to react with 117 g of hydrogen bromide, HBr. This result highlights the practical application of stoichiometry in chemistry. By understanding mole ratios and using balanced chemical equations, we can accurately predict the amounts of reactants needed for a complete reaction. This is crucial in various fields, from pharmaceutical research to industrial chemical production. Knowing the exact quantities of reactants ensures efficient reactions, minimizes waste, and maximizes the yield of desired products. So, guys, stoichiometry isn't just a theoretical concept; it's a powerful tool in the hands of chemists.

Key Takeaways

  • Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions.
  • A balanced chemical equation is essential for stoichiometric calculations as it provides the mole ratios between substances.
  • The molar mass is used to convert grams to moles and vice versa.
  • The stoichiometric ratio from the balanced equation is used to convert moles of one substance to moles of another.

Practice Problems

To solidify your understanding, let's try a few more practice problems. These problems will give you a chance to apply the concepts we've discussed and build your confidence in solving stoichiometry questions. Remember, practice makes perfect, especially in chemistry! So, grab a pen and paper, and let's get started.

  1. How many moles of oxygen gas (O₂) are needed to react completely with 4 moles of methane (CH₄) in the following reaction?

    CH₄ + 2 O₂ → CO₂ + 2 H₂O

  2. If you have 50 grams of sodium chloride (NaCl), how many moles do you have?

  3. In the reaction:

    N₂ + 3 H₂ → 2 NH₃

    How many moles of ammonia (NH₃) can be produced from 6 moles of hydrogen (H₂)?

These practice problems cover the key concepts we've discussed, from converting grams to moles to using stoichiometric ratios. Work through them step-by-step, and don't hesitate to refer back to the explanations and examples we've covered. The more you practice, the more comfortable you'll become with stoichiometry. It's like learning a new language; the more you use it, the more fluent you become.

Conclusion

Mastering stoichiometry is a fundamental skill in chemistry. By understanding how to convert grams to moles and using stoichiometric ratios from balanced chemical equations, you can solve a wide range of problems related to chemical reactions. This knowledge is not only essential for academic success but also for various real-world applications in fields like medicine, engineering, and environmental science. So, keep practicing, keep exploring, and keep asking questions. Chemistry, like any science, is a journey of discovery, and stoichiometry is a valuable tool on that path. Remember, guys, chemistry is all around us, and understanding it can help us understand the world a little better. Whether you're baking a cake or designing a new drug, the principles of stoichiometry are at play, guiding the proportions and ensuring the desired outcome. So, embrace the challenge, and enjoy the journey!