Question

A gas stream contains 18.0 mole% 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. (a) What is the flow rate of the gas stream leaving the condenser in mol/min? (Hint : First calculate the molar flow rate of the condensate and note that the rates at which C6H14 and N2 enter the unit must equal the total rates at which they leave in the two exit streams.) (b) What percentage of the hexane entering the condenser is recovered as a liquid?

Answer 1

A. 72.53 mol/min

B. 76.2%

Step-by-step explanation:

Please check attachment to see the diagram of the components I drew.

I first converted flow rate to a molar flow rate. I got 11.5mol/ min after converting. Check attachment

A.) When we create a balance on all these components

n1 + n2 + n3

n1 = n2 + 11.5

Creating a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.5mol/min(1.00)

When we put the first equation into the second one:

(n2+11.5mol/min)0.18 + n2(0.05) + 11.5mol/min(1.00)

= 0.18n2 + 2.07min/mol = 0.5n2 + 11.5mol/min

Collect like terms

0.18n2 - 0.05n2 = 11.5 - 2.07min/mol

= 0.13n2 = 9.43

n2 = 9.43/0.13

n2 = 72.53 mol/min

72.53 is the flow rate of the gas stream leaving the burner.

B.) n1 = n2 + 11.5mol/min

n1 = 72.53mol/min + 11.5mol/min

n1 = 84.93mol/min

Amount of hexane entering condenser

n1c6H14 = y1C6H14 = 0.18(84.03mol/min)

= 15.1mol/min

The percentage condensed

= (11.5mol/min)/(15.1mol/min) * 100

= 76.2% of hexane is recovered as liquid