Question

A mixture of bases can sometimes be the active ingredient in antacid tablets. If 0.4851 g of a mixture of Al(OH)₃ and Mg(OH)₂ is neutralized with 17.17 mL of 1.000 M HNO₃, what is the mass % of Al(OH)₃in the mixture?

Answer 1

Molar mass of Mg(OH)2 =58.32 g / mol

Molar mass of HNO3 = 63.01 g / mol

Molar mass of Al(OH)3 = 78.0 g / mol

Al(OH)3 + 3HNO3 --------> Al(NO3)3 +3H2O

1 mole Al hydroxide reacts with 3 moles HNO3

So, 78 g Al hydroxide reacts with=3*63.01 g HNO3

=189.03 g HNO3

Similarly,

Mg(OH)2 + 2HNO3 ----------> Mg(NO3)2 + 2H2O

1of magnesium hydroxide reacts with 2 moles HNO3
Or, 58.32 g magnesium hydroxide reacts with=2*63.01 g HNO3

=126.02 g HNO3

Let there is x g Al(OH)3 and( 0.4851 -x ) g Mg(OH)2
Total consumption of HNO3 = 0.01717 * 63.01 = 1.08 g HNO3

Therefore, 189.03/78*x + (0.4851-x)*126.02/58.32 = 1.08
2.4235x +2.161*(0. 4851-x) = 1.08
2.4235x+ 1.048 - 2.161x = 1.08
0.2625x = 0.032
x is 0.047/0.2625 = 0.121 g

therefore mass of Al(OH)3 = 0.121 g
mass of Mg(OH)2 is (0.4851-0.121)= 0.3641 g

% mass of Al(OH)3 is 0.121/0.4851 * 100 = 24.9 %