Question

Complete neutralization of a mixture of magnesium hydroxide and aluminum hydroxide is required 48.5 ml of 0.187M hydrochloric acid. The mixture of dry chloride salts from the reaction has the quantity of 0.420 g. What is the percentage by mass of magnesium hydroxide in the mixture of the above hydroxides for the reaction with HCl?

Answer 1

Solution:
Mg(OH)2 + 2HCl = MgCl2 + 2H2O (1)
2Al(OH)3 + 6HCl = 2AlCl3 + 3H2O (2)
Quantity of HCl: 0.187 M x 48.5 mL / 1000 mL = 0.0090695 mol
From which x – reacts with Mg(OH)2 , y - with Al(OH)3
So x + y = 0.0090695
Reaction (1): n (MgCl2) = x/2 mol; m (MgCl2) = x/2 ∙ 95.2110 g/mol
Reaction (2): n (AlCl3) = y/3 mol; m (AlCl3) = y/3 ∙ 133.3405 g/mol
x/2 ∙ 95.2110 + y/3 ∙ 133.3405 = 0.420

System:
x + y = 0.0090695
x/2 ∙ 95.2110 + y/3 ∙ 133.3405 = 0.420
y 0.003722
x 0.005347
Then:
m (Mg(OH)2) = x/2 ∙ 58.3197 g/mol = 0.155918 g
m (Al(OH)3) = y/3 ∙ 78.0036 g/mol = 0.096789 g
The sum: 0.252707 g
%mass Mg(OH)2 = 61.7 %
Answer: 61.7 %.