Answer to Question #317382 in Chemistry for Jjmoon

Question

In the reaction Mg(OH)2 + 2HCl → MgCl2 + 2H2 O, you were given 15

grams of Mg(OH)2 and 25 grams of HCl. Which of the two reactants is the

limiting reagent? How much MgCl2 will be produced in the reaction? (Mg(OH)2

= 58.32 g/mol, HCl = 36.46 g/mol, MgCl2

= 95.32 g/mol)

Answer 1

Solution:

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 / 36.46 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 reacts 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 produces:

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

Calculate the mass of MgCl₂:

The molar mass of MgCl₂ is 95.21 g/mol

Therefore,

Mass of MgCl₂ = [0.2572 mol MgCl₂] × [95.21 g MgCl₂ / 1 mol MgCl₂] = 24.488 g MgCl₂ = 24.5 g MgCl₂

Mass of MgCl₂ = 24.5 g

Answers:

The limiting reagent is Mg(OH)₂

24.5 grams of MgCl₂ will be produced in the reaction