Answer:

Explanation:

To find the total weight of the precipitate when pure aluminum sulfate is treated with excess barium hydroxide solution, you need to understand the chemical reaction that occurs.

When aluminum sulfate reacts with barium hydroxide, it forms barium sulfate and aluminum hydroxide as precipitates:

Al2(SO4)3 + 6Ba(OH)2 → 3BaSO4 + 2Al(OH)3

Now, let's calculate the molar mass of the compounds involved:

Molar mass of BaSO4 (barium sulfate) = 137.33 g/mol + 32.07 g/mol + 4 * 16.00 g/mol = 233.33 g/mol

Molar mass of Al(OH)3 (aluminum hydroxide) = 26.98 g/mol + 3 * (16.00 g/mol + 1.01 g/mol) = 78.01 g/mol

Now, let's calculate the molar mass of aluminum ions (Al³⁺):

Molar mass of Al³⁺ = 26.98 g/mol

Given that you have 0.170 g of aluminum ions in the solution, you can calculate the number of moles of aluminum ions:

Number of moles of Al³⁺ = Mass / Molar mass = 0.170 g / 26.98 g/mol = 0.006298 moles

Now, according to the balanced chemical equation, the ratio of aluminum hydroxide to aluminum ions is 2:3. So, the number of moles of aluminum hydroxide formed will be:

Number of moles of Al(OH)3 = (2/3) * 0.006298 moles = 0.0041987 moles

Now, let's find the mass of aluminum hydroxide formed:

Mass of Al(OH)3 = Number of moles * Molar mass = 0.0041987 moles * 78.01 g/mol = 0.3269 g

So, the total weight of the precipitate, which includes both barium sulfate and aluminum hydroxide, is the sum of their masses:

Total weight of the precipitate = Mass of BaSO4 + Mass of Al(OH)3

Total weight of the precipitate = 0.170 g (BaSO4) + 0.3269 g (Al(OH)3) = 0.4969 g

Rounded to two decimal places, this is approximately 0.50 g. Therefore, the correct answer is option (1) 0.5 g.