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A mixture of toluene and air is passed through a cooler where some toluene is condensed....

A mixture of toluene and air is passed through a cooler where some toluene is condensed. One thousand cubic feet of gases enter the cooler per hour at 100 oC and 100 mm Hg gauge. The partial pressure of the toluene in this mixture is 300 mm Hg. If 720 ft3 / hr of gaseous mixture comes out of the cooler at 50 oC and barometric pressure of 760 mm Hg. Calculate the pounds of toluene removed per hour in the cooler.

Solutions

Expert Solution

flow rate of gases containing toluene and air at 100deg.c= 1000ft3/hr= 1000*28.32 L/hr (1ft3=28.32L) = 28320 L/hr

pressure= 100mm (g), absolute pressure= 100+760 mm Hg= 860mm Hg= 860/760 atm (1atm=760mmHg)= 1.13 atm

moles of mixture at these conditions = PV/RT= 1.13*28320/(0.0821*373)= 1045 moles/hr

moles of gases at the exit conditions similarly= 720*28.32L/hr*760/760/(0.0821*323)= 769 moles/hr

only toluene gets condensed, hence moles of toluene condensed= 1045-769= 276 moles/hr

molar mass of toluene (C6H5CH3)=92, mass of toluene condensed = moles of toluene*molar mass= 276*92 gm=25392 gm/hr= 25392/1000=25.392 kg/hr

since 1 lb=0.4535Kg

25.392 kg/hr= 25.392/0.4535 lb/ hr=55.99 lb/hr

queen_honey_blossom answered 2 years ago
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