air is bubbled through a drum of liquid hexane at a rate of 0.200 kmol / min. the gas stream leaving the drum contains 10.0 mole's hexane vapor. air may be considered insoluble in liquid hexane. use an integral balance to estirmate the time reguired to vaporize in min to vaporize 10.0 m of the liquid. if the total moles of the products 0.222 kmol / min and the specific gravity of liquid hexane is 0.659 and the accumulation term equals -76.45 kmol ch4

Complete Question

Air is bubbled through a drum of liquid hexane at a rate of 0.200 kmol / min. the gas stream leaving the drum contains 10.0% mole's hexane vapor. air may be considered insoluble in liquid hexane. use an integral balance to estirmate the time reguired to vaporize in min to vaporize 10.0 m^3  of the liquid.if the total moles of the products 0.222 kmol / min and the specific gravity of liquid hexane is 0.659 and the accumulation term equals -76.45 kmol ch4

The  value is  

Explanation:

From the question we are told that

 The rate at which the air is bubbled is  

  The  rate at which the gas stream is leaving the drum is  

 The  percentage of hexane contained in the gas leaving the drum  

 The volume of liquid hexane to be vaporized is

  The  specific gravity of liquid hexane is

Generally the density of the liquid is mathematically evaluated as

Here is the density of water with value

So  

The mass of the liquid is mathematically represented as

The number of moles of liquid hexane is mathematically represented as

Here  M is the molar mass of hexane with value  

  So

The rate at which hexane liquid is vaporized is mathematically represented as

Generally the time required to vaporize the the liquid(i.e  n) is mathematically represented as

=>    

=>      

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.

i think a, hope it .

explanation: