Question

*Please show how you got the calculations.

In a reactor scenario, 20.000 g Fe3O4 was reacted, at 1450 K, with NO having an initial pressure of 3.000 atm. At equilibrium, the mass of the solid-phase product (Fe2O3) was calculated to be 8.0742 g. 2Fe3O4(s) + 2NO(g) <=> 3Fe2O3(s) + N2O(g)

1. Calculate the mass of each gas-phase species at equilibrium.

2. Calculate the mass of Fe3O4 remaining at equilibrium from the mass of Fe2O3 calculated to exist at equilibrium.

3. Calculate the total mass of all species at equilibrium and prove that it equals the initial total mass of the reactants, to the .1 g place.

Answer 1

20.000 g Fe3O4 was reacted, at 1450 K, with NO having an initial pressure of 3.000 atm. At equilibrium, the mass of the solid-phase product (Fe2O3) was calculated to be 8.0742 g.

2Fe3O4(s) + 2NO(g) <=> 3Fe2O3(s) + N2O(g)

Solution:

Mass of Fe3O4 = 20g so moles of Fe3O4 = mass / molecular wt

= 20 / 231.5 = 0.086 moles

Mass of Fe2O3 = 8.0472 / 159.6 = 0.0504 moles

1. Calculate the mass of each gas-phase species at equilibrium.

         2Fe3O4(s)     +        2NO(g)      <=>        3Fe2O3(s)    +   N2O(g)

Initial       0.086 moles             3 atm                               0                    0

Change       -2x                         -2x                                  3x                   x

Equilibrium     0.086-2x                                                     0.0504  

So 3x = 0.0504

x = 0.0504 / 3 = 0.0168 moles

so mass of N2O at equilibrium = moles X molecular wt = 0.0168 X 44 = 0.739 grams

Assuming that the volume of container is 1L

The initial moles of NO = PV / RT = 3 X 1 / 1450 X 0.0821 = 0.025 moles

So moles of NO left at equilibrium = 0.025 – 2x = 0.025 - 2X 0.0168 = 0.0086 moles

                            

1. Calculate the mass of Fe3O4 remaining at equilibrium from the mass of Fe2O3 calculated to exist at equilibrium.

   Moles of Fe3O4 left = 0.086 - 0.0168 X 2 = 0.0524 moles

   0.0524 moles = Mass / molwt

   Mass = 0.0524 X 231.5 = 12.13 g

2. Calculate the total mass of all species at equilibrium and prove that it equals the initial total mass of the reactants, to the .1 g place.

   Solution

   Initial Mass of NO = Moles X molecular wt = 0.025 X 30 = 0.75 g

   Initial mass of Fe3O4 = 20g

   So total mass = 20.7g

   After equilibrium

   Mass of NO left = moles X molecular wt = 0.0086 X 30 = 0.2g

   Mass of Fe3O4 left = moles X molecular wt = 0.0524 X 231.5 = 12.1 g

   Mass of Fe2O3 formed moles X molecular wt= 8.0 g

   Mass of N2O formed = moles X molecular wt= 0.7 grams

Mass = 0.2 + 12.1 + 8.0 + 0.7 =   21 approx.