Answer to Question #331533 in General Chemistry for Ronette

Question

In the reaction Mg(OH)²+2HCl--->MgCl²+2H²0 you were given 15 grams of Mg(OH)² and 25 grams of HCl.Which of the two reactants is the limiting reagents? How much MgCl² will be produced in the reaction?

Answer 1

Solution:

The molar mass of Mg(OH)₂ is 58.32 g/mol

The molar mass of HCl is 36.458 g/mol

Calculate moles of each reagent:

Moles of Mg(OH)₂ = [15 g Mg(OH)₂] × [1 mol Mg(OH)₂ / 58.32 g Mg(OH)₂] = 0.2572 mol Mg(OH)₂

Moles of HCl = [25 g HCl] × [1 mol HCl / 36.458 g HCl] = 0.6857 mol HCl

Balanced chemical equation:

Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

According to stoichiometry:

1 mol of Mg(OH)₂ reacts with 2 mol of HCl

Thus, 0.2572 mol of Mg(OH)₂ react with:

[0.2572 mol Mg(OH)₂] × [2 mol HCl / 1 mol Mg(OH)₂] = 0.5144 mol HCl

However, initially there is 0.6857 mol of HCl (according to the task)

Therefore, Mg(OH)₂ acts as limiting reagent and HCl is excess reagent

According to stoichiometry:

1 mol of Mg(OH)₂ produces 1 mol of MgCl₂

Thus, 0.2572 mol of Mg(OH)₂ produce:

[0.2572 mol Mg(OH)₂] × [1 mol MgCl₂ / 1 mol Mg(OH)₂] = 0.2572 mol MgCl₂

The molar mass of MgCl₂ is 95.211 g/mol

Therefore,

Mass of MgCl₂ = [0.2572 mol MgCl₂] × [95.211 g MgCl₂ / 1 mol MgCl₂] = 24.488 g MgCl₂ = 24.5 g MgCl₂

Mass of MgCl₂ = 24.5 g

Answer:

The limiting reagent is Mg(OH)₂

24.5 grams of MgCl₂ will be produced in the reaction