Question

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?

Answer 1

Answer:

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.