Question

A gas stream contains 18.0 mol% hexane and the remainder nitrogen. the stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. the hexane mole fraction in the gas stream leaving the condenser is 0.0500. liquid hexane condensate is recovered at a rate of 1.50 l/min

Answer 1

The question relates to the condensation of a hexane and nitrogen gas mixture. Calculations involving hexane mole fraction changes, liquid recovery rates, and concepts from gas laws and solution chemistry are necessary to provide answers to this query.

Step-by-step explanation:

The question deals with a gas stream containing hexane and nitrogen which undergoes condensation to remove some of the hexane. The hexane mole fraction in the departing gas stream, the hexane condensate recovery rate, and possibly calculations involving mole fractions, partial pressures, and the ideal gas law are inquired. Key principles involved are the behavior of gas mixtures, condensation processes, and the properties of ideal solutions when temperature adjustments lead to phase change and composition alterations.

Since the primary focus is on the change in mole fraction of hexane after condensation and how it relates to the amount of liquid hexane recovered, we delve into gas laws and solution chemistry to provide precise answers. Calculations may include using Raoult's Law for the partial pressures of heptane and octane in a mixed solution or determining partial pressures of components in natural gas based on their mole percentages.

Answer 2

There is no question specified for this problem. Based on my understanding, there are 5 possible questions that would come up. Once you know them, the whole problem is solved. They are:

a.) The amount of entering gas stream (denoted as A)
b.) The amount of exiting gas stream (denoted as B)
c.) The amount of condensate (denoted as C)
d.) Mole fraction of liquid hexane (x,h)
e.) Mole fraction of liquid nitrogen (x,n)

In order to solve this, we apply mass balance.
Overall Mass Balance: A = B + C
Hexane Balance: 0.18A = 0.05B + 1.5
Nitrogen Balance: 0.82A = 0.95B + (C - 1.50)

Solving the equations simultaneously:
0.18(B+C) = 0.05B + 1.50
C = (1.5 - 0.13B)/0.18

0.82[B + ((1.5 - 0.13B)/0.18)] = 0.95 + ((1.5 - 0.13B)/0.18) - 1.5
Solving for B,
B = 1 L/min
Then,
C = (1.5 - 0.13(1))/0.18 = 7.611 L/min
A = 1 L/min + 7.611 L/min = 8.611 L/min
x,h = 1.5/C = 1.5/7.611 = 0.197
x,n = 1 - 0.197 = 0.803

As a summary, the possible answers to the possible questions are:
a.) 8.611 L/min entering gas stream
b.) 1 L/min exiting gas stream
c.) 7.611 L/min condensate
d.) The condensate is 19.7 mol% hexane
e.) The condensate is 80.3 mol% nitrogen