Question

consider the following reaction of ammonium carbonate decomposing upon heating according to the equation below:
(NH4)2 Co3 (s) -> 2NH3 (g) + CO2 (g) + l2o (g)

a. calc the total value of (l) gas produces at 22°c and 1.02 atm when 11.83 of ammonium carbonate is reacted

b. calculate the partial pressure (atm) and make fraction of the ammonia gas (NH3) in the product mixture.

Answer 1

a) (NH4)2CO3 moles = mass / ( Molar mass of (NH4)2CO3)

             = ( 11.83g) / ( 96g/mol) = 0.1232 mol

Total gas moles produced = 4 moles ( 2NH3 + 1CO2 + 1H2O = 4 gas moles) per 1 mole of ( NH4)2CO3

hence when 0.1232 mol of (NH4)2CO3 reacted we get 4 x 0.1232 = 0.493 mol

b) NH3 moles produced = 2 x (NH4)2CO3 moles = 2 x 0.1232 = 0.2464

mol fraction of NH3 = moles of NH3 / total gas moles = ( 0.2464/0.493) = 0.5

Partial pressure of NH3 = total pressure x NH3 mol fraction

                = 1.02 x 0.5 = 0.51 atm