16 45

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What percentage of the antacid is active ingredient?

What percentage of the antacid is active ingredient?
The antacid in question (with an active ingredient of CaCO3), upon treatment with excess hydrochloric acid, a 0.2308 g sample gives 16.45 mL of CO2 at 21.00oC and 98.45KPa. What percentage of the antacid is active ingredient? Though it is not a perfect system- 80.31% yield. How do I go about doing this with steps?

determine moles of CO2 produced from PV=nRT
n= PV/RT
n= 0.9716 atm x 0.01645 L/(0.082 x 294.15)
n= 6.626E-4 moles CO2

CaCO3 + 2 HCl -----> CO2 + CaCl2 + H2O
1 mole CO2 is produced per mole of CaCO3, therefore sample contained 6.626E-4 moles CaCO3
6.626E-4 moles CaCO3 x 100.09 g/mole CaCO3=0.0663 g CaCO3 (theo.)
0.0663/.8031 yield= 0.0826 g CaCO3 in sample

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