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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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michele10 michele10
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11 years ago
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%)
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#1
11 years ago
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
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