An antacid tablet weighs 2.10 grams. It requires 45.67 mL of 0.105 M of HCI solution to react completely with the carbonate present in the tablet. Determine the mass of CaCO3 present in the tablet?

≅ 0.240 grams (3 sig. figs.)

Explanation:

Rxn:   CaCO₃  +          2HCl              => CaCl₂ + H₂O + CO₂

Given:    ?g          45.67ml(0.105M)

                            = 0.04567L x 0.105 mole/L

                            =  0.0048 mole HCl

Rxn ratio for CaCO₃ to HCl is 1:2

∴ moles CaCO₃ consumed = 1/2 of moles HCl used

=> 1/2(0.0048)mole CaCO₃ used = 0.0024 mole CaCO₃

mass CaCO₃ = 0.0024 mole CaCO₃  x  100.09 grams CaCO₃/mole CaCO₃

= 0.23998 grams CaCO₃ (calculator answer)

≅ 0.240 grams (3 sig. figs.)