Calculate $BaSO_4$ Mass From $BaCl_2$ And $Na_2SO_4$

Hey guys! Today, we're diving into a classic chemistry problem: figuring out how much barium sulfate ($BaSO_4$) forms when barium chloride ($BaCl_2$) reacts with sodium sulfate ($Na_2SO_4$). This is a common type of problem in stoichiometry, and it’s super useful for understanding chemical reactions and quantitative analysis. So, let's break it down step-by-step.

Understanding the Reaction

First, let’s write down the balanced chemical equation. This helps us see the ratio in which the reactants combine and the products form:

$BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)$

From this equation, we can see that one mole of barium chloride ($BaCl_2$) reacts with one mole of sodium sulfate ($Na_2SO_4$) to produce one mole of barium sulfate ($BaSO_4$) and two moles of sodium chloride ($NaCl$). The key here is the 1:1 stoichiometric ratio between $BaCl_2$ and $BaSO_4$. This ratio is crucial for calculating the mass of $BaSO_4$ formed.

Initial Conditions

We're given that we have 200.0 mL of a 0.10 M $BaCl_2$ solution. The concentration of a solution tells us how many moles of solute (in this case, $BaCl_2$) are dissolved in a liter of solution. Molarity (M) is defined as moles of solute per liter of solution (mol/L). So, a 0.10 M $BaCl_2$ solution contains 0.10 moles of $BaCl_2$ in every liter of solution.

We're also told that $Na_2SO_4$ is in excess. This means we have more than enough $Na_2SO_4$ to react completely with all the $BaCl_2$. Therefore, $BaCl_2$ is the limiting reactant – it will determine how much $BaSO_4$ is produced. Excess reagent ensures that the reaction proceeds to completion, maximizing the yield of the desired product.

Step-by-Step Calculation

Here’s how we can calculate the mass of $BaSO_4$ formed:

Step 1: Calculate Moles of $BaCl_2$

To find the number of moles of $BaCl_2$, we use the formula:

Moles = Molarity × Volume (in liters)

First, we need to convert the volume of the $BaCl_2$ solution from milliliters to liters:

Volume = 200.0 mL = 200.0 / 1000 L = 0.200 L

Now, we can calculate the moles of $BaCl_2$:

Moles of $BaCl_2$ = 0.10 M × 0.200 L = 0.020 moles

Step 2: Determine Moles of $BaSO_4$

Since the stoichiometric ratio between $BaCl_2$ and $BaSO_4$ is 1:1, the number of moles of $BaSO_4$ produced is equal to the number of moles of $BaCl_2$ reacted:

Moles of $BaSO_4$ = Moles of $BaCl_2$ = 0.020 moles

Step 3: Calculate Mass of $BaSO_4$

To find the mass of $BaSO_4$, we use the formula:

Mass = Moles × Molar Mass

The molar mass of $BaSO_4$ can be calculated by adding the atomic masses of each element in the compound:

• Barium (Ba): 137.33 g/mol
• Sulfur (S): 32.07 g/mol
• Oxygen (O): 16.00 g/mol (and we have 4 of them)

Molar mass of $BaSO_4$ = 137.33 + 32.07 + (4 × 16.00) = 137.33 + 32.07 + 64.00 = 233.40 g/mol

Now we can calculate the mass of $BaSO_4$:

Mass of $BaSO_4$ = 0.020 moles × 233.40 g/mol = 4.668 g

Therefore, the mass of $BaSO_4$ formed when 200.0 mL of 0.10 M $BaCl_2$ reacts with excess $Na_2SO_4$ is approximately 4.668 grams.

Additional Considerations

Significant Figures

In our calculations, we should pay attention to significant figures. The initial volume (200.0 mL) and concentration (0.10 M) both have two significant figures. Therefore, our final answer should also be rounded to two significant figures:

Mass of $BaSO_4$ ≈ 4.7 g

Practical Implications

The precipitation of $BaSO_4$ is often used in gravimetric analysis, a quantitative laboratory technique used to determine the concentration of an analyte (the substance being measured) through the measurement of its mass. $BaSO_4$ is highly insoluble in water, which makes it an excellent choice for gravimetric analysis of barium or sulfate ions. The precipitate is carefully filtered, dried, and then weighed to determine the amount of the analyte in the original sample.

Common Mistakes to Avoid

Forgetting to Convert Units

Always ensure that your units are consistent. For example, volume should be in liters when using molarity to calculate moles.

Incorrect Stoichiometric Ratios

Double-check the balanced chemical equation to ensure you are using the correct stoichiometric ratios. In this case, the 1:1 ratio between $BaCl_2$ and $BaSO_4$ made the calculation straightforward, but other reactions may have different ratios.

Using the Wrong Molar Mass

Make sure you calculate the molar mass of the compound correctly. A small mistake here can lead to a significant error in your final answer.

Not Considering Limiting Reactant

Always identify the limiting reactant when one of the reactants is in excess. This ensures that you are using the correct amount of reactant to calculate the product yield.

Conclusion

So, there you have it! By following these steps, you can easily calculate the mass of $BaSO_4$ formed in this reaction. Remember to pay attention to the stoichiometry of the reaction, use the correct units, and avoid common mistakes. With a bit of practice, you'll become a pro at solving stoichiometry problems!

Understanding how to calculate the mass of a precipitate like barium sulfate is not only a fundamental skill in chemistry but also has practical applications in various analytical techniques. Whether you're working in a lab or just studying chemistry, mastering these calculations will give you a solid foundation in quantitative analysis. Keep practicing, and you'll be amazed at how quickly you improve. Happy calculating, everyone!