Question

A mixture of bases can sometimes be the active ingredient in antacid tablets. If 0.4884 g of a mixture of Al(OH)3 and Mg(OH)2 is neutralized with 17.12 mL of 1.000 M HNO3, what is the mass % of Al(OH)3 in the mixture?

Answer 1

54.7%

Step-by-step explanation:

The reaction of neutralization will be:

Al(OH)₃ + Mg(OH)₂ + 5HNO₃ → Al(NO₃)₃ + Mg(NO₃)₂ + 5H₂O

The number of moles of HNO₃ added is the concentration multiplied by the volume (17.12 mL = 0.01712 L)

n = 1*0.01712 = 0.01712 mol

For the stoichiometry:

1 mol of Al(OH)₃ ---------- 5 moles of HNO₃

x ----------------------------------0.01712 mol

By a simple direct three rule:

5x = 0.01712

x = 3.424x10⁻³ mol of Al(OH)₃

For the stoichiometry, the number of moles of Mg(OH)₂ will also be 3.424x10⁻³.

The molar masses are:

Al(OH)₃: 27 g/mol of Al + 3*16 g/mol of O + 3*1 g/mol of H = 78 g/mol

Mg(OH)₂: 24 g/mol of Mg + 2*16 g/mol of O + 2*1 g/mol of H = 58 g/mol

The mass is the molar mass multiplied by the number of moles:

Al(OH)₃: 78*3.424x10⁻³ = 0.2671 g

Mg(OH)₂: 58*3.424x10⁻³ = 0.1986 g

The mass % of Al(OH)₃ in the mixture is its mass divides by the mass of the mixture, multiplied by 100%:

(0.2671/0.4884)x100% = 54.7%