dissolving 4.69g of an impure sample of calcium carbonate in hydrochloric acid p

Question

dissolving 4.69g of an impure sample of calcium carbonate in hydrochloric acid produced 1.07 L of carbon dioxide (measured at 20.0 degrees Celcius and 737mmHg). Calculate the percent by mass of calcium carbonate in the sample.

Explanation / Answer

CaCO3 + 2HCl ---> CaCl2 + CO2 + H2O so 1 mole of CaCO3 gives one mole of CO2 1.07L of CO2 at 20 degree C =20+273K =293 K 737mm Hg= 737/760 atm=0.9697 atm (as760 mm Hg = 1 atm ) n = PV/RT = (0.9697atm) x (1.07 L) / ((0.0821 Latm/moleK) x (293K)) =0.0431 mol 1 mol of CaCO3=100.0875 g therefore 0.0431mol= 4.31 g % of CaCO3= (4.31/4.69)x100=91.978%

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