Al (OH) 3 + 3 HCl = AlCl3 + 3 H2O

Some brands of antacid tablets have used aluminum hydroxide to neutralize excess acid, as shown above. How many grams of HCl can a tablet with 0.880 g of Al (OH) 3 consume? How much h2O?

1) 1.235 g.

2) 0.61 g.

Explanation:

From the balanced equation:

Al (OH) ₃ + 3HCl → AlCl₃ + 3H₂O.

1.0 mol of Al (OH) ₃ reacts with 3.0 moles of HCl to produce 1.0 mol of AlCl₃ and 3.0 moles of H₂O.

1) How many grams of HCl can a tablet with 0.880 g of Al (OH) ₃ consume?

To calculate the amount of HCl needed to consume 0.880 g of Al (OH) ₃, we need to calculate the no. of moles of Al (OH) ₃:

no. of moles of Al (OH) ₃ = mass/molar mass = (0.880 g) / (78.0 g/mol) = 1.13 x 10⁻² mol.

∵ Every 1.0 mol of Al (OH) ₃ needs 3.0 moles of HCl to be consumed.

∴ 1.13 x 10⁻² mol of Al (OH) ₃ needs (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of HCl.

The no. of grams of HCl = no. of moles of HCl x molar mass of HCl = (3.385 x 10⁻² mol) (36.5 g/mol) = 1.235 g.

2) How much H₂O?

∵ Every 1.0 mol of Al (OH) ₃ produces 3.0 moles of H₂O.

∴ 1.13 x 10⁻² mol of Al (OH) ₃ produces (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of H₂O.

The no. of grams of H₂O = no. of moles of H₂O x molar mass of H₂O = (3.385 x 10⁻² mol) (18.0 g/mol) = 0.6092 g ≅ 0.61 g.