Question

`CO_2` gas, in the dry state, may be produced by heating calcium carbonate. `CaCO_3 (s)DeltaCaO(s) + CO_2(g)`. What volume of `CO_2`, collected dry at 55 C and a pressure of 774 torr, is produced by complete thermal decomposition of 10.0 g of `CaCO_3`?

Answer 1

3.12 Liters if Dry Pressure is 656mmHg (Water Vapor Pressure - 118mmHg- is subtracted from 744mmHg), or 2.75 Liters if Dry Pressure is 744 mmHg. Calculation that follows is based upon 656mmHg dry pressure (=0.863Atm). Substitute 1.02Atm if 744mmHg is dry pressure.

Explanation:

`10gramsCaCO_3(s) => CaO(s) + (?)O_2(g)`

`"moles" CaCO_3 = ((10g)/(100(g/(mol))))` = `0.10"mole"`

From reaction ratios
=> moles `CaCO_3(s)` consumed = moles `O_2(g)` produced
= 0.10 mole `(O_2(g))`****

If a 'dry' volume is needed (`VP_(H_2O)=118mmHg@55^oC)`, then the water vapor pressure at `55^oC` is subtracted from the given 774mmHg pressure.
=> Dry Pressure = 774mmHg - 118mmHg = 656mmHg
656mmHg = `(656mmHg)/(((760mmHg)/(Atm))` = 0.863Atm

Temperature = `55^oC + 273` = 328K

`R = (0.08206(((L)(Atm))/((mol)(K)))`

Using the Ideal Gas Law => `PV = nRT => V_(dry) = (nRT)/(P)`

`V` = `((0.10mol)(0.08206((L)(Atm))/((mol)(K)))(328K))/(0.863Atm) = 3.12L`

OR => If dry pressure is 774torr = 1.02Atm => `V_(dry) = 2.75L`