Question

8. When a calcium carbonate tablet is ingested, it dissolves by the reaction of stomach acid, which contains hydrochloric acid. The unbalanced equation for this reaction is CaCO,(s) + HCl(aq) -----> CaCl(aq) + H2O(l) + CO.(g). Normal stomach acid contains 5.50% hydrochloric acid. If a particular individual required 13 calcium carbonate tablets to completely neutralize their stomach acid, how many grams of stomach acid reacted? Each calcium carbonate tablet contains 500. mg of the active ingredient.

Answer 1

CaCO3 + 2 HCl --> CaCl2 + CO2 +H2O

1mole CaCO3 neutralizes 2 mols HCl

we have molar mass of CaCO3 = [1*40.08+ 1*12.01+1*16]g/mol = 100.09 g/mol

mass of CaCO3 consumed= 13* 500 mg= 6.5 g

moles of CaCO3 consumed = mass/molar mass= 6.5 g/100.09 g/mol= 0.0649 mols

hence acid neutralized = 2* 0.0649 mols = 0.1298 moles

molar mass of HCL= 36.46 g /mol

mass of HCl present = moles of HCl*molar mass of HCl

= 0.1298 moles* 36.46 g/mol = 4.7325 g

mass of stomach acid present = 4.7325 g

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