Respuesta :
Explanation:
The given data is as follows.
Mass flow rate of mixture = 1368 kg/hr
[tex]N_{2}[/tex] in feed = 40 mole%
This means that [tex]H_{2}[/tex] in feed = (100 - 40)% = 60%
We assume that there are 100 total moles/hr of gas [tex](N_{2} + H_{2})[/tex] in feed stream.
Hence, calculate the total mass flow rate as follows.
40 moles/hr of N_{2}/hr (28 g/mol of [tex]N_{2}[/tex]) + 60 moles/hr of [tex]H_{2}/hr[/tex] (2 g/mol of [tex]H_{2}[/tex])
[tex]40 \times 28 g/hr + 60 \times 2 g/hr[/tex]
= 1120 g/hr + 120 g/hr
= 1240 g/hr
= [tex]\frac{1240}{1000}[/tex] (as 1 kg = 1000 g)
= 1.240 kg/hr
Now, we will calculate mol/hr in the actual feed stream as follows.
[tex]\frac{100 mol/hr}{1.240 kg/hr} \times 1368 kg/hr[/tex]
= 110322.58 moles/hr
It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of [tex]N_{2}[/tex] into the reactor as follows.
[tex]0.4 \times 110322.58 moles/hr[/tex]
= 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 [tex]N_{2}[/tex] into the reactor as follows.
[tex]0.028 kg \times 44129.03 mol/hr[/tex]
= 1235.612 kg/hr
Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.
