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

Step-by-step 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 * 28 g/hr + 60 * 2 g/hr`

= 1120 g/hr + 120 g/hr

= 1240 g/hr

=
`(1240)/(1000)` (as 1 kg = 1000 g)

= 1.240 kg/hr

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

`(100 mol/hr)/(1.240 kg/hr) * 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 * 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 * 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.