In a combustion furnace, 2094 standard ft3 per hour
of natural gas (methane) is burned with 6% excess air.
how many standard ft3 of air are drawn from outside
per hour by the fan that supplies the air?

Explanation:

The chemical reaction is as follows.

It is given that 2094 . And, it is known that 1 = 127133

Hence, convert 2094 into as follows.

                  =

As ideal gas equation is PV = nRT. So, calculate the number of moles as follows.

                   n =

                      =

                       = 0.673 mol/sec

According to the stoichiometry of the given reaction, 1 mol of methane reacts with 2 mol of oxygen.

So, 1 mol = 2 mol = 1.346 mol/s

Hence, air required theoretically = = 6.4095 mol/s.

Since, 6% of excess air is being supplied. Therefore, total air supplied will be calculated as follows.

                Total air supplied =

                                              = 6.794 mol/s

Now, calculate the volume using ideal gas law equation as follows.

                            PV = nRT

                           V = 0.166229

Converting calculated volume into as follows.

                  1 = 127133

So,        0.166229 =  

                                        = 21133.191

Thus, we can conclude that 21133.191 of air are drawn from outside  per hour by the fan that supplies the air.

h let me think

yes it is balanced