If one regular antacid tablet contains 500 mg of solid CaCO₃ (100 g/mol), how many mL of 1.0 M stomach acid (HCl) could it neutralize? A)5 mL B)10 mL C)50 mL D)100 mL E)15 mL

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Quizplus · Accepted Answer

The reaction between CaCO3 and HCl is CaCO3 + 2HCl → CaCl2 + CO2 + H2O. Since one mole of CaCO3 reacts with two moles of HCl, and one tablet contains 5 moles of CaCO3 (500 mg / 100 g/mol = 5 mmol = 0.005 mol), it can neutralize 0.01 moles of HCl. Since the concentration of HCl is 1.0 M, which means 1 mole per liter, 0.01 moles of HCl is contained in 10 mL of solution (0.01 moles / 1.0 M = 0.01 L = 10 mL).

If One Regular Antacid Tablet Contains 500 Mg of Solid