Using the equation below, determine the volume (in liters) of CO2 that is produced at 12,500 Pa and 298 K from a 12.0 g sample of CaCO3.

Question

1. Using the equation below, determine the volume (in liters) of CO2 that is produced at 12,500 Pa and 298 K from a 12.0 g sample of CaCO3.

1 CaCO3 + 2 HCl → 1 CaCl2 + 1 CO2 + 1 H2O

Answer 1

First determine the number of moles of CaCO₃ using the molar mass.

The molar mass of CaCO₃ is:

Ca:x 1 x 40.078 = 40.078

C x 1 x 12.011 = 12.011

O x 3 x 15.999 = 47.997

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100.086 g/mol

12g/100.086 g/mol = 0.1199 moles of CaCO₃

Using the chemical reaction equation we see that for every 1 mole of CaCO₃ there is 1 mole of CO₂ produced so for 0.1199 moles of CaCO₃ there are 0.1199 moles of CO₂ produced.

Now, using the Ideal Gas Law (PV = nRT) we can solve for V by dividing both sides by P making:

V = nRT/P

We can use the value of R at 8.314462 L•kPa/(K•mol) but we'll need to convert the pressure to kPa by dividing it by 1000 making the pressure 12.5 KPa

Plugging in the numbers we get:

V = nRT/P

V = (0.1199)(8.314462)(298)/12.5 = 23.8 L