Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 l of carbon dioxide at 20.0 Â°c and 791 mmhg. Calculate the percent by mass of calcium carbonate in the sample.

calcium carbonate was dissolved completely and the amount of carbon dioxide released was proportional to the amount of calcium carbonate.
the sample did not contain any other compound that released or reacted with carbon dioxide.

[tex]PV = nRT[/tex]

[tex]n = \frac{PV}{RT}[/tex]

[tex]P = 791 mmHg = \frac{791 mmHg}{760 mmHg} x 1 atm = 1.040789 atm.[/tex]

[tex]R = 8.314 \frac{J}{K*mol} = 0.082 \frac{L*atm}{K*mol}[/tex]

[tex]T = 20^{0}C + 273^{0} C = 293 K[/tex]

[tex]n = \frac{1.040789 atm * 1.14 L}{0.082 \frac{L * atm}{K * mol} * 293 K }[/tex]

[tex]n = 0.0493839 mol.[/tex]

given the equation of the reaction:

[tex]CaCO_{3} + 2HCL = CaCl_{2} + CO_{2} + H_{2}O[/tex]

from assumption:

amount carbon dioxide = amount of calcium carbonate

[tex]n(CaCO_{3} ) = n(CO_{2} )[/tex]

reacting mass ( m ) = Molar Mass ( M ) * Amount ( n )

[tex]m(CaCO_{3} ) = n(CaCO_{3} ) * M(CaCO_{3} )[/tex]

m = 0.04938397 [tex]mol[/tex] * 100 [tex]\frac{g}{mol}[/tex] = 4.93839 [tex]g[/tex]

percentage by mass of CaC[tex]O_{3}[/tex] = [tex]\frac{mass of pure }{mass of impure}*100 = \frac{4.93839g}{5.28g}*100 = 93.5802%[/tex].