Question

Calculate the maximum volume in mL of 0.20 M HCl that a tablet containing 338 mg Al(OH)3 and 489 mg Mg(OH)2 would be expected to neutralize. Assume complete neutralization.

Answer 1

148.85 mL

Step-by-step explanation:

The reactions that take place are:

• 3HCl + Al(OH)₃ → AlCl₃ + 3H₂O

• 2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O

Now we calculate how many HCl moles would react with 338 mg of Al(OH)₃. First we convert Al(OH)₃ mg into mmol, using its molar mass:

• 338 mg Al(OH)₃ ÷ 78 mg/mmol = 4.33 mmol Al(OH)₃

Then we convert mmol Al(OH)₃ into mmol HCl:

• 4.33 mmol Al(OH)₃ *
  `(3mmolHCl)/(1mmolAl(OH)_(3))` = 13 mmol HCl

Now we do the same calculations for Mg(OH)₂, using its molar mass:

• 489 mg Mg(OH)₂ ÷ 58.32 mg/mmol = 8.38 mmol Mg(OH)₂

And convert into mmol HCl:

• 8.38 mmol Mg(OH)₂ *
  `(2mmolHCl)/(1mmolMg(OH)_(2))` = 16.77 mmol HCl

Now we add the reacting mmoles of HCl together

• 13 mmol HCl + 16.77 mmol HCl = 29.77 mmol HCl

Finally we calculate the volume of a 0.20 M solution that contains 29.77 mmoles of HCl:

• Molarity = mmol HCl / mL

• 0.20 M = 29.77 mmol HCl / V

• V = 148.85 mL