Calculate the amount of the salt produced by the reaction between 8.4 g magnesium hydroxide and 80 mL of 1.0 M sulfuric acid.

Question

Also calculate amount of excess reactant remaining from the reaction

Answer 1

Balance the reaction to get mole ratios: Mg(OH)₂ + H₂SO₄ --> MgSO₄ + 2 H₂O. Stoichiometry:

[8.4g Mg(OH)₂] * [1 mol Mg(OH)₂ /58.32g Mg(OH)₂] * [1 mol MgSO₄ /1 mol Mg(OH)₂] = 0.14₄ mol MgSO₄.

80 mL = 0.080 L; (0.080 L H₂SO₄) * (M = 1.0 mol/L) * (1 mol MgSO₄ /1 mol H₂SO₄) = 0.080 mol MgSO₄.

Sulfuric acid is limiting, (0.080 mol MgSO₄) * (120.36g MgSO₄ /1 mol MgSO₄) = 9.6g MgSO₄ salt.

Excess (Δmole = 0.064 mol MgSO₄) * (1 mol Mg(OH)₂ /1 mol MgSO₄) * [58.32g Mg(OH)₂ /1 mol Mg(OH)₂] = 3.7g Mg(OH)₂ excess reactant.