A. Calculate the mass of the salt that can be produced from 10.0 g of magnesium chloride reacting with sulfuric acid. Include the percent yield of the salt if 11.5 g was actually produced.

Question

A. Calculate the mass of the salt that can be produced from 10.0 g of magnesium chloride

reacting with sulfuric acid. Include the percent yield of the salt if 11.5 g was actually produced.

B. 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. (Hint- First find the Limiting Reactant)

C.Calculate amount of excess reactant remaining from the reaction in the clue above.

Answer 1

A) Balance the reaction to get mole ratios: MgCl₂ + H₂SO₄ --> MgSO₄ + 2 HCl. Stoichiometry:

(10.0g MgCl₂) * (1 mole MgCl₂ /95.205g MgCl₂) * (1 mole MgSO₄ /1 mole MgCl₂) * (120.36g MgSO₄ /1 mole MgSO₄) = 12.6 g MgSO₄ salt, theoretically.

The percent yield is 100 * actual / theoretical = 100 * (11.5 / 12.6₄) = 91.0 %.

B) 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) = 9.6 g MgSO₄ salt.

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