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. N2 C6H14(v) (5.00 mole%) N2 C6H14(v) (18.0 mole%) Liquid condensate 1.50 L C6H14 (l)/min CONDENSER (a) What is the flow rate of the gas stream leaving the condenser in mol/min

Answer 1

the flow rate of the gas stream leaving the condenser is 71.9 moles/min

Explanation:

Given the data in the question and the figure below;

N2 BALANCE

(100% - 18%) × ( mole rate in ) = ( 100% - 5%) × ( mole rate out)

0.82 × ( mole rate in ) = 0.95 × ( mole rate out)

mole rate in = 0.95 × ( mole rate out) / 0.82

mole rate in = 1.1585365853 × ( mole rate out)

now;

HEXANE BALANCE

0.18 × ( mole rate in ) = 0.0500 × ( mole rate out) + condensate --- equ 1

but condensate = 1.5 L/min × ( density of hexane ) × 1/molar mass of hexane

we know that;

density of hexane is 0.6548 g/mL

and molar mass of hexane is 86.18 g/mol

so,

condensate = 1.5 L/min × ( 0.6548 g/mL × 1000 mL/L ) × ( 1/86.18 g/mol)

condensate = 11.3970758876

now lets substitute into equation 1

0.18 × ( mole rate in ) = 0.0500 × ( mole rate out) + condensate

⇒ 0.18 × ( 1.1585365853 × ( mole rate out) ) = 0.0500 × ( mole rate out) + 11.3970758876

⇒ 0.208536585354(mole rate out) = 0.0500( mole rate out) + 11.3970758876

⇒ 0.208536585354(mole rate out) - 0.0500( mole rate out) = 11.3970758876

⇒ 0.158536585354(mole rate out) = 11.3970758876

mole rate out = 11.3970758876 / 0.158536585354

mole rate out = 71.889247 ≈ 71.9 moles/min

Therefore, the flow rate of the gas stream leaving the condenser is 71.9 moles/min