Alloy 5 g of aluminum and 5 g of sulfur. The resulting alloy was treated with water.

Alloy 5 g of aluminum and 5 g of sulfur. The resulting alloy was treated with water. Calculate the mass of the solid precipitate.

1. Let us write down the equations of the proceeding reactions:

2Al + 3S = Al2S3;

Al2S3 + 6H2O = 2Al (OH) 3 + 3H2S;

2.Calculate the chemical quantities of portions of aluminum and sulfur:

n (Al) = m (Al): M (Al) = 5: 27 = 0.185 mol;

n (S) = m (S): M (Al) = 5: 32 = 0.156 mol;

3.Let’s find the amount of sulfide by the amount of aluminum:

n (Al2S3) = n (Al): 2 = 0.185: 2 = 0.0925 mol;

4. find the amount of sulfide through the amount of sulfur:

n (Al2S3) = 2 * n (S): 3 = 2 * 0.156: 3 = 0.104 mol, for further calculations we will use the chemical amount found through aluminum, since it is in short supply;

5. find the amount of aluminum hydroxide:

n (Al (OH) 3) = 2 * n (Al2S3) = 2 * 0.0925 = 0.185 mol;

6. find the mass of the sediment:

m (Al (OH) 3) = n (Al (OH) 3) * M (Al (OH) 3);

M (Al (OH) 3) = 27 + 3 * 1 + 3 * 16 = 78 g / mol;

m (Al (OH) 3) = 0.185 * 78 = 14.43 g.

Answer: 14.43 g.