Consider the following reaction, calculate the mass of NH4NO3 needed to produce a total equilibrium pressure of 2.03 atm at 500 oC in a 1.5 L container.

Question

NH4NO3 (s)= N2O (g) + 2H2O (g)

Answer 1

                            NH4NO3 (s)= N2O (g) + 2H2O (g)

at t=0                      0                  0               0

at equilibrium            0                  p               2p          ( only gases can have pressure)

 

 

according to question at eq. total pressure is 2.03 atm.  that means,  p+2p= 2.03 atm    or 3p= 2.03 atm  or p= 2/3 atm.

           now , pressure of N2O (g) = p= 2/3 atm.    

                    pressure of H2O (g)= 2p= 2* 2/3 atm.

     temp = 500+273= 773K

 

as per ideal gas equation, PV=nRT or n= PV/RT

 

for number of moles of N2O (g)= (2/3 atm)(1.5 lit)/0.0821L atm per K per mol)*(773K)= 0.0158 moles

     number of moles of H2O( g)= (2*2/3 atm)(1.5 lit)/.0821L atm per K per mol)*(773K)=0.0315 moles

 

NOW if 1 MOLE 0F NH4NO3 (s) is taken it will produces total 3 moles of product.

 moles ,

 

          NH4NO3 (s)= N2O (g) + 2H2O (g)

at t=0       1              0             0

at eq         1-x           x            2x

 

  

 

now as we solved , x= 0.0158 moles(moles of N2O (g)

                            moles of NH4NO3 (s) at eq= 1-x or 1-0.0158= 0.9842 moles .  

mol wt of NH4NO3= 80.04336

 

now moles= mass/mol wt    or mass= moles*mol wt of NH4NO3 = 0.9842* 80.04336 = 78.778 g