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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.

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Answer:

148.85 mL

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 * [tex]\frac{3mmolHCl}{1mmolAl(OH)_{3}}[/tex] =  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 *  [tex]\frac{2mmolHCl}{1mmolMg(OH)_{2}}[/tex] =  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

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