Answer:
91.38 %
Explanation:
First we use the following formula to calculate the number of moles of HCl and NaOH contained in the solutions:
molar concentration = number of moles / solution volume
number of moles = molar concentration × solution volume
number of moles of HCl = 0.2050 × 100 = 20.5 mmoles
number of moles of NaOH = 0.1020 × 19.85 = 2.02 mmoles
reaction between NaOH and HCl:
NaOH + HCl → NaCl + H₂O
If 2.02 mmoles of NaOH were used for neutralization of HCl left from the reaction, which are 2.02 mmoles of HCl, it means that the HCl reacted with Mg(OH)₂:
20.5 - 2.02 = 18.48 mmoles of HCl
2 HCl + Mg(OH)₂ = MgCl₂ + 2 H₂O
From the chemical reaction we devise the following reasoning:
if 2 mmoles of HCl react with 1 mmole of Mg(OH)₂
then 18.48 mmoles HCl react with X mmoles of Mg(OH)₂
X = (18.48 × 1) / 2 = 9.24 mmoles of Mg(OH)₂
now the mass of Mg(OH)₂:
mass = number of moles × molecular weight
mass of Mg(OH)₂ = 9.24 × 58.3 = 538.7 mg = 0.5387 g
Now the percent by mass of magnesium hydroxide Mg(OH)₂ in the impure sample (purity):
purity = (mass of pure Mg(OH)₂ / mass of impure sample) × 100
purity = (0.5387 / 0.5895) × 100
purity = 91.38 %
