Amixture of a and b is capable of being ignited only if the mole percent of a is 6 %. a mixture containing 9.0 mole% a in b flowing at a rate of 800 kg/h is to be diluted with pure b to reduce a concentration to the lower flammability limit. calculate the required flow rate of b in mol/h and the percent by mass of oz in the product gas. (note: b may be taken to consist of 21 mole% o2 and 79% nz and to have an average molecular weight of 29.0.) ma = 16.04 g/mol.

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

             Weight =

                          = 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

                     = 28.74 kmol/hr

Now, applying component balance as follows.

                    = 43.11 kmol/hr

                          = 43.11 - 28.74

                                      = 14.37 kmol/hr

So, mass flow rate of pure (B), is =

                                                    = 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

             Molecular weight of product stream =

                                  =

                                  = 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of into the product stream is as follows.

                      =

                      = 272.31 kg/hr

Therefore, calculate the mass % of into the stream as follows.

                     = 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of in the product gas is 22.38%.

answer: they produce more slow than unicellular organisms.