Answer:
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%
