Question

Ammonia is produced by the following reaction in laboratory: 2NH4Cl(s) + Ca(OH)2(s) → CaCl2(s) + 2H2O(l) + 2NH3(g) In one reaction, 4.0 kg of ammonium chloride is heated at 529 °C in a 5.0-L vessel. The pressure of NH3 is 0.16 bar after 8.0 minutes.

a) What is the average rate of NH3 production in mol/min during the 8-minute interval?

b) How many moles of ammonium chloride consumed in the 8-minute interval?

Answer 1

a)The pressure of NH3 = 0.16bar

temperature = 529 C = 529+273 =802K

volume = 5.0L

From the reaction NH3 is the only gas present in reaction mixture .

Thus using ideal gas PV = nRT we can calculate the number of moles of NH3 formed.

Thus number of moles of NH3 = PV/RT

= 0.16 atm x 5.0L / 0.0821L.atm/K.mol x 802K

= 0.01215 mol

Average Rate of formation of NH3 = increase in concentration / time taken

= 0.1215 mol/ 8 min

= 1.519x10^-3 mol/min

b) From the stoichiometric equation

2 moles of NH4Cl reacts to produce 2 moles of NH3

Thus moles of NH3 produced = moles of NH4Cl consumed

Thus moles of NH4Cl consumed = 0.01215 mol