#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#?

1 Answer
Doc048
May 28, 2017

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#

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