Question

1. A antacid tablet is 85% calcium carbonate by mass. The tablet is fully dissolved into 180.0mL of 0.170 M HCL, leaving excess HCL. A volume of 19.00 mL of 0.245 M KOH is needed to neutralize the remaining acid.
   1. What is the full molecular equations for both neutralization reactions occurring?
   2. What is the original mass of the tablet used?

Answer 1

(a.) First reaction between calcium carbonate CaCO₃ and HCl

CaCO₃ (s) + 2 HCl (aq) CaCl₂ (aq) + CO₂ (g) + H₂O (l)

Second reaction between KOH and HCl

KOH (aq) + HCl (aq) KCl (aq) + H₂O (l)

(b.) Concentration HCl = 0.170 M

volume HCl = 180.0 mL = 0.1800 L

Total moles HCl added = (Concentration HCl) * (volume HCl in Liter)

Total moles HCl added = (0.170 M) * (0.1800 L)

Total moles HCl added = 0.0306 mol

moles KOH added = (concentration KOH) * (volume KOH in Liter)

moles KOH added = (0.245 M) * (0.01900 L)

moles KOH added = 0.004655 mol

moles HCl reacted with KOH = moles KOH added

moles HCl reacted with KOH = 0.004655 mol

moles HCl reacted with CaCO₃ = (total moles HCl) - (moles HCl reacted with KOH)

moles HCl reacted with CaCO₃ = (0.0306 mol) - (0.004655 mol)

moles HCl reacted with CaCO₃ = 0.025945 mol

moles CaCO₃ reacted = (moles HCl reacted with CaCO₃) * (1 mole CaCO₃ / 2 moles HCl)

moles CaCO₃ reacted = (0.025945 mol) * (1 / 2)

moles CaCO₃ reacted = (0.025945 mol) * (0.5)

moles CaCO₃ reacted = 0.01297 mol

mass CaCO₃ reacted = (moles CaCO₃ reacted) * (molar mass CaCO₃)

mass CaCO₃ reacted = (0.01297 mol) * (100.09 g/mol)

mass CaCO₃ reacted = 1.30 g

mass of tablet = (mass CaCO₃) / (mass fraction CaCO₃)

mass of tablet = (1.30 g) / (0.85)

mass of tablet = 1.53 g