Amixture 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?

54.7%

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%

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >