Question

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

Answer 1

Answer:

Percentage by mass of calcium carbonate in the sample is 93.58%.

Explanation:

Assumptions:

• 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.

$PV = nRT$

$n = \frac{PV}{RT}$

$P = 791 mmHg = \frac{791 mmHg}{760 mmHg} x 1 atm = 1.040789 atm.$

$R = 8.314 \frac{J}{K*mol} = 0.082 \frac{L*atm}{K*mol}$

$T = 20^{0}C + 273^{0} C = 293 K$

$n = \frac{1.040789 atm * 1.14 L}{0.082 \frac{L * atm}{K * mol} * 293 K }$

$n = 0.0493839 mol.$

given the equation of the reaction:

$CaCO_{3} + 2HCL = CaCl_{2} + CO_{2} + H_{2}O$

from assumption:

amount carbon dioxide = amount of calcium carbonate

$n(CaCO_{3} ) = n(CO_{2} )$

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

$m(CaCO_{3} ) = n(CaCO_{3} ) * M(CaCO_{3} )$

m = 0.04938397 $mol$ * 100 $\frac{g}{mol}$ =  4.93839 $g$

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