You're asking about a chemical reaction, but you haven't provided a complete reaction. To determine the grams of aluminum hydroxide (Al(OH)₃) obtained from 15.7 grams of sulfide, we need to know:

1. What specific sulfide compound is reacting? There are many sulfides (e.g., sodium sulfide (Na₂S), hydrogen sulfide (H₂S), iron sulfide (FeS)).

2. What is the other reactant in the reaction? The amount of aluminum hydroxide produced depends on the reaction's stoichiometry, and we need to know the other reactant and its amount.

Here's a general example of how to solve this type of problem:

Let's assume the reaction is:

3 Na₂S + 2 AlCl₃ → 2 Al(OH)₃ + 6 NaCl

To calculate the grams of Al(OH)₃:

1. Convert grams of Na₂S to moles:

- Find the molar mass of Na₂S (2 x 23 g/mol Na + 32 g/mol S = 78 g/mol).

- Divide the given mass of Na₂S by its molar mass: 15.7 g Na₂S / 78 g/mol Na₂S = 0.201 mol Na₂S

2. Use the stoichiometry to find moles of Al(OH)₃:

- The balanced equation shows that 3 moles of Na₂S produce 2 moles of Al(OH)₃.

- Set up a proportion: (0.201 mol Na₂S / 3 mol Na₂S) = (x mol Al(OH)₃ / 2 mol Al(OH)₃)

- Solve for x: x = 0.134 mol Al(OH)₃

3. Convert moles of Al(OH)₃ to grams:

- Find the molar mass of Al(OH)₃ (27 g/mol Al + 3 x 16 g/mol O + 3 x 1 g/mol H = 78 g/mol).

- Multiply the moles of Al(OH)₃ by its molar mass: 0.134 mol Al(OH)₃ x 78 g/mol Al(OH)₃ = 10.5 g Al(OH)₃

Therefore, in this example, 10.5 grams of aluminum hydroxide would be obtained.

Please provide the complete reaction and the other reactant, and I can help you calculate the grams of aluminum hydroxide.