Barium Chloride Calculation: Precipitating Sulfate Ions

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Hey guys! Let's dive into a chemistry problem. This one involves figuring out how much barium chloride you need to make all those sulfate ions in a solution go poof and form a solid precipitate. We'll break it down step-by-step, making sure it's super clear and easy to follow. Get ready to flex those chemistry muscles! So, the task is to calculate the mass of barium chloride (BaCl₂) required to completely precipitate sulfate ions (SO₄²⁻) from a potassium sulfate (K₂SO₄) solution. The solution has a mass of 870 g and a mass fraction of potassium sulfate of 0.1 (or 10%). This kind of problem often appears in chemistry quizzes, and understanding the approach can be very helpful for exam success.

Understanding the Chemistry Behind the Problem

First off, let's talk about what's happening chemically. When barium chloride (BaCl₂) is added to a solution containing sulfate ions (SO₄²⁻), a precipitation reaction occurs. This means that an insoluble solid, barium sulfate (BaSO₄), is formed. The chemical equation for this reaction is:

BaCl₂(aq) + K₂SO₄(aq) -> BaSO₄(s) + 2KCl(aq)

In this equation:

  • BaCl₂(aq) represents aqueous (dissolved in water) barium chloride.
  • K₂SO₄(aq) represents aqueous potassium sulfate.
  • BaSO₄(s) represents solid (precipitated) barium sulfate.
  • KCl(aq) represents aqueous potassium chloride. This is a spectator ion in terms of our calculation, and we don't care about it. It just hangs around in the solution.

The key thing here is that one mole of BaCl₂ reacts with one mole of K₂SO₄ to produce one mole of BaSO₄. Because the question is about the mass of barium chloride, we need to know how much potassium sulfate we're dealing with.

Now, the mass fraction of potassium sulfate of 0.1 means that 10% of the solution's mass is made up of K₂SO₄. Since our solution weighs 870 grams, we can easily calculate the mass of the potassium sulfate present. This is the first step! Remember, understanding the chemical reaction is the cornerstone of solving this problem.

We will take the mass fraction of potassium sulfate in the solution to find the mass of potassium sulfate in the solution. Then, we will find the number of moles of potassium sulfate in the solution. We will use the stoichiometry of the reaction to find the number of moles of barium chloride required for the reaction, and we will finally convert the number of moles of barium chloride to grams, which is the final answer.

Step-by-Step Calculation Guide

Alright, let's roll up our sleeves and crunch some numbers! Follow along, and you'll become a precipitation pro in no time.

Step 1: Calculate the mass of potassium sulfate (K₂SO₄).

  • Mass of solution = 870 g
  • Mass fraction of K₂SO₄ = 0.1
  • Mass of K₂SO₄ = Mass of solution × Mass fraction of K₂SO₄ = 870 g × 0.1 = 87 g

So, there are 87 grams of potassium sulfate in the solution. Cool, huh?

Step 2: Calculate the number of moles of potassium sulfate (K₂SO₄).

To do this, we'll need the molar mass of K₂SO₄. You can find this by adding up the atomic masses of all the atoms in the molecule, which you'll typically find on the periodic table.

  • Molar mass of K = 39.1 g/mol (approximately) × 2 = 78.2 g/mol
  • Molar mass of S = 32.1 g/mol (approximately)
  • Molar mass of O = 16.0 g/mol (approximately) × 4 = 64.0 g/mol
  • Molar mass of K₂SO₄ = 78.2 g/mol + 32.1 g/mol + 64.0 g/mol = 174.3 g/mol

Now we can calculate the number of moles:

Moles of K₂SO₄ = Mass of K₂SO₄ / Molar mass of K₂SO₄ = 87 g / 174.3 g/mol ≈ 0.499 mol

Step 3: Determine the moles of barium chloride (BaCl₂) needed.

From the balanced chemical equation, we know that 1 mole of BaCl₂ reacts with 1 mole of K₂SO₄. Therefore, the number of moles of BaCl₂ needed is equal to the number of moles of K₂SO₄.

Moles of BaCl₂ = Moles of K₂SO₄ ≈ 0.499 mol

Step 4: Calculate the mass of barium chloride (BaCl₂).

We need the molar mass of BaCl₂ to do this. Again, we add up the atomic masses from the periodic table.

  • Molar mass of Ba = 137.3 g/mol (approximately)
  • Molar mass of Cl = 35.5 g/mol (approximately) × 2 = 71.0 g/mol
  • Molar mass of BaCl₂ = 137.3 g/mol + 71.0 g/mol = 208.3 g/mol

Now, calculate the mass:

Mass of BaCl₂ = Moles of BaCl₂ × Molar mass of BaCl₂ ≈ 0.499 mol × 208.3 g/mol ≈ 104.0 g

Therefore, approximately 104.0 grams of barium chloride are required to precipitate the sulfate ions from the solution.

Important Considerations and Tips

  • Accuracy: Always pay attention to significant figures in your calculations. They show how precise your measurements are. The more data you use, the better and more accurate your answer will be.
  • Units: Make sure your units are consistent throughout the calculations (grams, moles, etc.).
  • Molar Masses: Double-check the molar masses you use. Small errors here can lead to big errors in your final answer. The best method for finding this is to refer to a periodic table.
  • Complete Precipitation: The problem asks for the mass needed to completely precipitate the sulfate ions. This implies that we need enough BaCl₂ to react with all the K₂SO₄ present.
  • Practice: The best way to get good at these types of problems is to practice! Try working through similar examples to solidify your understanding.

Conclusion: You Got This!

There you have it! We've successfully calculated the mass of barium chloride needed. It may seem like a lot to take in at first, but with a step-by-step approach and a good understanding of the chemistry involved, these problems become much more manageable. Keep practicing, and you'll be acing these questions in no time. Chemistry can be fun, guys!

If you have any questions or want to try another problem, drop a comment below! Happy calculating!