What percentage of the antacid is active ingredient?

Question

The antacid in question (with an active ingredient of Ca(HCO3)2), upon treatment with excess hydrochloric acid, a 0.433 g sample gives 88.15 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 + 2CO2 + 2H2O

Atomic weights: Ca=40 H=1 C=12 O=16 Ca(HCO3)2=162

Let the antacid be called A. Let Ca(HCO3)2, the active ingredient, be called CB. 25.00C=298.00K

88.15mLCO2/0.433gA x 273K/298K x 1molCO2/22,400mLCO2 x 1molCB/2molCO2 x 162gCB/1molCB x 100% = 67.4% CB