Question

Pure nitrogen (N2) and pure hydrogen (H2) are fed to a mixer. The product stream has 40.0% mole nitrogen and the balance hydrogen. The product flow rate is 1,368 kg/hr. Calculate the feed rate of pure nitrogen to the mixer in kg/hr.

Answer 1

Explanation:

The given data is as follows.

        Mass flow rate of mixture = 1368 kg/hr

      $N_{2}$ in feed = 40 mole%

This means that $H_{2}$ in feed = (100 - 40)% = 60%

We assume that there are 100 total moles/hr of gas $(N_{2} + H_{2})$ in feed stream.

Hence, calculate the total mass flow rate as follows.

           40 moles/hr of N_{2}/hr (28 g/mol of $N_{2}$) + 60 moles/hr of $H_{2}/hr$ (2 g/mol of $H_{2}$)

                  $40 \times 28 g/hr + 60 \times 2 g/hr$    

                  = 1120 g/hr + 120 g/hr

                  = 1240 g/hr

                  = $\frac{1240}{1000}$              (as 1 kg = 1000 g)

                  = 1.240 kg/hr

Now, we will calculate mol/hr in the actual feed stream as follows.

                 $\frac{100 mol/hr}{1.240 kg/hr} \times 1368 kg/hr$

                   = 110322.58 moles/hr

It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of $N_{2}$ into the reactor as follows.

                       $0.4 \times 110322.58 moles/hr$

                      = 44129.03 mol/hr

As 1 mole of nitrogen has 28 g/mol of mass or 0.028 kg.

Therefore, calculate the rate flow of $N_{2}$ into the reactor as follows.

                       $0.028 kg \times 44129.03 mol/hr$

                         = 1235.612 kg/hr

Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.