Respuesta :
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 = [tex]0.09 \times 16.04 + 0.91 \times 29[/tex]
= 27.8336 g/mol
And, molar flow rate of air and mixture is calculated as follows.
[tex]\frac{800}{27.8336}[/tex]
= 28.74 kmol/hr
Now, applying component balance as follows.
[tex]0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}[/tex]
[tex]F_{p}[/tex] = 43.11 kmol/hr
[tex]F_{A} + F_{B} = F_{p}[/tex]
[tex]F_{B}[/tex] = 43.11 - 28.74
= 14.37 kmol/hr
So, mass flow rate of pure (B), is [tex]F_{B}[/tex] = [tex]14.37 \times 29[/tex]
= 416.73 kg/hr
According to the product stream, 6 mol% A and 94 mol% B is there.
Molecular weight of product stream = [tex]Mol. weight \times 43.11 kmol/hr[/tex]
= [tex]0.06 \times 16.04 + 0.94 \times 29[/tex]
= 28.22 g/mol
Mass of product stream = 1216.67 kg/hr
Hence, mole of [tex]O_{2}[/tex] into the product stream is as follows.
[tex]0.21 \times 0.94 \times 43.11[/tex]
= [tex]8.5099 kmol/hr \times 329 g/mol[/tex]
= 272.31 kg/hr
Therefore, calculate the mass % of [tex]O_{2}[/tex] into the stream as follows.
[tex]\frac{272.31}{1216.67} \times 100[/tex]
= 22.38%
Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of [tex]O_{2}[/tex] in the product gas is 22.38%.
