A chemist reacts 35 g of MgCl, with excess KOH according to the reaction below. How much Mg(OH)2 would be
produced if the reaction goes to completion?
MgCl2 + 2KOH – Mg(OH)2 + 2KCI

Mass = 21.58 g

Explanation:

Given data:

Mass of MgCl₂ = 35 g

Mass of Mg(OH)₂ produced = ?

Solution:

Chemical equation:

MgCl₂ + 2KOH    →     Mg(OH)₂ + 2KCl

Number of moles of MgCl₂:

Number of moles = mass/molar mass

Number of moles = 35 g/ 95.2 g/mol

Number of moles = 0.37 mol

now we will compare the moles of MgCl₂ and Mg(OH)₂

                MgCl₂         :        Mg(OH)₂

                   1               :            1

                 0.37           :        0.37

Mass of  Mg(OH)₂ :

Mass = number of moles × molar mass

Mass = 0.37 mol × 58.32 g/mol

Mass = 21.58 g

the ratio of the conjugate base to the acid required to attain the desired ph.