Question

Stomach acid is approximately 0.10 M HCl. How many mL of stomach acid can be neutralized by one regular antacid tablet that contains 500 mg of solid CaCO3 (100.09 g/mol)?

Answer 1

Answer:

100 mL

Explanation:

The reaction that takes place is:

• CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

First we convert 500 mg of CaCO₃ into mmoles, using its molar mass:

• 500 mg ÷ 100 mg/mmol = 5 mmol CaCO₃

Then we convert 5 mmoles of CaCO₃ into HCl mmoles, using the stoichiometric coefficients of the balanced reaction:

• 5 mmol CaCO₃ * $\frac{2mmolHCl}{1mmolCaCO_3}$ = 10 mmol HCl

Finally we calculate the volume of a 0.10 M HCl solution (such as stomach acid) that would contain 10 mmoles:

• 10 mmol / 0.10 M = 100 mL