Question

An important chemical reaction in the manfacture of Portland cement is the high temperature decomposition of calcium carbonate to give calcium oxide and carbondioxde

CaCo3(s) -------Co2(g0 +Cao(s) . Suppose a 1.25 g sample of CaC03 is decomposed by heating. How many millimeters of Co2 gas will be evolved if the volume will be measured at 745 torr and 25 degree centigrade?

Answer 1

ANSWER:

The given decomposition reaction of calcium carbonate to give calcium oxide and carbon dioxide is:

CaCO₃(s) CO₂(g) +CaO(s)

Here, we have

mass of CaCO₃ = 1.25 g

molar mass of CaCO₃ = 100.087 g/mol

number of moles of CaCO₃ = (1.25 g)/(100.087 g/mol)

= 0.0125 mol

For CO₂ :

pressure, P = 745 torr

temperature, T = 25 ^oC = 298.15 K

gas constant, R = 62.364 L torr mol^-1 K^-1

We see that,

for 1 mole of CaCO₃, we are getting = 1 mole of CO₂

so, for 0.0125 moles of CaCO₃, we will get = 0.0125 moles of CO₂

from ideal gas equation,

PV = nRT

745 torr x V = 0.0125 mol x 62.364 L torr mol^-1 K^-1 x 298.15 K

V = 0.31198 L = 311.98 mL

Hence, 311.98 milliliter of CO₂ gas will be evolved if the volume will be measured at 745 Torr and 25 ^oC.

{Note: in place of millimeter there should be milliliter}