The antacid in question (with an active ingredient of Ca(HCO3)2), upon treatment with excess hydrochloric acid, a 0.413 g sample gives 92.85 mL of CO2 at 25.00oC and 1.000 atm. What percentage of the antacid is active ingredient? Assume a perfect system (i.e. the yield is 100%)

Answer 1

Ca(HCO3)2 + 2HCl ==> CaCl2 + 2H2O + 2CO2

Use PV = nRT. Substitute and solve for n.
mols x molar mass = grams CO2.
Convert g CO2 to g Ca(HCO3)2
(g Ca(HCO3)2/mass sample)*100 = % Ca(HCO3)2

Answer 2

dont you have to convert mols CO2 to mols Ca(HCO3)2 first?

Answer 3

dont you have to convert mols CO2 to mols Ca(HCO3)2 first? then convert to get grams?