Please help, any and all is very much appreciated!
1 answer:
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Explanation:
The given data is as follows.
Mass flow rate of mixture = 1368 kg/hr
in feed = 40 mole%
This means that in feed = (100 - 40)% = 60%
We assume that there are 100 total moles/hr of gas in feed stream.
Hence, calculate the total mass flow rate as follows.
40 moles/hr of N_{2}/hr (28 g/mol of ) + 60 moles/hr of
(2 g/mol of
)
= 1120 g/hr + 120 g/hr
= 1240 g/hr
= (as 1 kg = 1000 g)
= 1.240 kg/hr
Now, we will calculate mol/hr in the actual feed stream as follows.
= 110322.58 moles/hr
It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of into the reactor as follows.
= 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 into the reactor as follows.
= 1235.612 kg/hr
Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.