2.34 g of sample of sulphuric acid was diluted with water and sulphate was precipitated with barium chloride solution . the precipitate weighed 5.573 g .calculate the percentage of sulfate radical in sulphuric acid

To determine the percentage of the sulfate radical in the sulfuric acid sample, we need to follow a series of logical steps involving stoichiometry and some basic chemistry principles. Let's break it down step by step.

Understanding the Reaction

When sulfuric acid (H₂SO₄) is diluted and treated with barium chloride (BaCl₂), a precipitation reaction occurs, forming barium sulfate (BaSO₄). The balanced chemical equation for this reaction is:

• H₂SO₄ + BaCl₂ → BaSO₄ (s) + 2 HCl

In this reaction, one mole of sulfuric acid produces one mole of barium sulfate. The key here is to find out how much sulfate is present in the original sulfuric acid sample based on the weight of the precipitate formed.

Calculating Moles of Barium Sulfate

The first step is to calculate the number of moles of barium sulfate formed from the precipitate. The molar mass of barium sulfate (BaSO₄) can be calculated as follows:

• Molar mass of Ba = 137.33 g/mol
• Molar mass of S = 32.07 g/mol
• Molar mass of O = 16.00 g/mol (4 oxygen atoms)

So, the total molar mass of BaSO₄ is:

137.33 + 32.07 + (4 × 16.00) = 233.39 g/mol

Now, using the weight of the precipitate (5.573 g), we can find the number of moles of BaSO₄:

Number of moles of BaSO₄ = mass / molar mass = 5.573 g / 233.39 g/mol ≈ 0.0239 moles

Determining Moles of Sulfate

Since the reaction shows that one mole of H₂SO₄ produces one mole of BaSO₄, the moles of sulfate (SO₄²⁻) in the sulfuric acid will also be 0.0239 moles.

Calculating Mass of Sulfate

Next, we need to find the mass of the sulfate radical. The molar mass of the sulfate ion (SO₄²⁻) is:

• Molar mass of S = 32.07 g/mol
• Molar mass of O = 16.00 g/mol (4 oxygen atoms)

Thus, the total molar mass of SO₄²⁻ is:

32.07 + (4 × 16.00) = 96.07 g/mol

Now, we can calculate the mass of sulfate in the sample:

Mass of sulfate = moles × molar mass = 0.0239 moles × 96.07 g/mol ≈ 2.295 g

Calculating Percentage of Sulfate in Sulfuric Acid

Finally, to find the percentage of the sulfate radical in the original sulfuric acid sample, we use the formula:

Percentage of sulfate = (mass of sulfate / mass of sulfuric acid sample) × 100

Substituting the values we have:

Percentage of sulfate = (2.295 g / 2.34 g) × 100 ≈ 98.09%

Final Thoughts

Thus, the percentage of the sulfate radical in the sulfuric acid sample is approximately 98.09%. This calculation illustrates the relationship between the precipitate formed and the original acid, showcasing the principles of stoichiometry and chemical reactions effectively.