A gas stream (A) of 15.0 mol% H2, and the balance N2, is to be mixed with another gas stream (B) containing 40.0 mol% H2, and the balance N2, to produce 100 kg/h of a 25 mol% H2, and the
balance N2 gas stream (C).
(a) Draw and fully label a flowchart of the mixing process.
(b) Calculate the average molecular weight of the product stream (C).
(c) Calculate the molar flow rates of the product stream (C) in kmol/h.
(d) Calculate the required flow rates of the feed mixtures A and B in kmol/h.

A gas stream and other points are discussed below in details.

Explanation:

The method flow chart is displayed above.

B)Mole fraction of H2 in stream C=25%=0.25

Mole fraction of N2 in steams C=1-Mole fraction of H2=1-0.25=0.75

Average molecular weight of stream C=Mole fraction of H2*Molecular weight of H2+ Mole fraction of N2*Molecular weight of N2=0.25*2 kg/kmol+0.75*28 kg/kmol=21.5 kg/kmol

C) Mass flow rate of product C=100 kg/h

Molar flow rate of product C=Mass flow rate /Molecular weight=100 kg/h/21.5 kg/kmol=4.6512 kmol/h

D) Apply overall mole balance

Mole in=Mole out

Mole in=Mole of A+Mole of B=A+B

Let mole of A=A kmol/h

Mole of B=B kmol/h

Mole out=Mole of C stream=4.6512 kmol/h

A+B=4.6512 eq.1

Apply hydrogen mole balance

Mole of H2 in=Mole of H2 in A stream+ Mole of H2 in B stream=0.15 A+0.4 B

Mole of H2 out=Mole of H2 in C stream=0.25*4.6512=1.1628 kmol/h

1.1628 =0.15 A+0.4 B eq.2

Multiply eq.1 by 0.15

0.69768=0.15 A+0.15 B eq.3

Subtract eq.3 from eq.2

0.25 B=0.46512

B=1.86048 kmol/h

Substitute in eq1.

A=4.6512-1.86048=2.79042 kmol/h

Molar flow rate of A=2.79042 kmol/h

A molar flow rate of B=1.86048 kmol/h

answer:

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explanation: