[jokerboy111]

I'm completely lost on this problem...its not HW, but it has been bothering me:

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 ATM and 298K is needed to completely burn 1.00 L of C6H14 (density is 0.660 g/mL)  to yield CO2 and H2O?


Do I make a balanced equation or somehow find the moles of oxygen, how does one go step by step considering each given of the equation? I really have no clue how to start/solve it, I just know that the answer in the key is 8490 L.

Thanks alot!

[Yggdrasil]

Do I make a balanced equation or somehow find the moles of oxygen, how does one go step by step considering each given of the equation?

Starting with a balanced chemical reaction is always a good idea.  Once you have a balanced chemical reaction for the combustion of C₆H₁₄, you can figure out the number of moles of oxygen that you need.

[jokerboy111]

Ok so I got the balanced equation to be

2C₆H₁₄ + 19O₂   -->  12CO₂ + 14H₂O

And I thought I could find the mole ratio by setting up the given ratio equal to the ratio equation:

0.2095= 19moles                moles of air = 71.6921   
       19moles + X moles air 


Am I even going in the right direction? Please help me, thanks...

[Borek]

Close, but no luck yet.

First - 19 moles O₂ per 2 moles of your hydrocarbon. So far you only know volume of liquid, not number of moles.

Second - your mole ratio is wrong, although you are moving in the right direction. X is not number of moles of air, but number of moles of gasesother then oxygen. You may assume it is all nitrogen if it helps.

[jokerboy111]

Ok so from what youre saying, i thought I could solve the fraction by:

19 mol O₂ / X moles Air = 0.2095

      X moles Air = 90.692 mol

Then what?

[Borek]

Calculate number of moles of hydrocarbon.

[jokerboy111]

Ok, I honestly have no idea what to do..

[Borek]

What is mass of hydrocarbon in question? What is its molar mass?

[jokerboy111]

Ok, the molar mass of the hydrocarbon is 86.2 g/mol.

So, is this right?

       ( .660 g   )(1 mol hydrocarbon)( 1.0 L used) (1000mL)             
            mL                 86.2 g                              ( 1 L )

    = 7.65661 mol hydrocarbon  ?

I have a feeling this lead me nowhere

[Yggdrasil]

Ok, the molar mass of the hydrocarbon is 86.2 g/mol.

So, is this right?

       ( .660 g   )(1 mol hydrocarbon)( 1.0 L used) (1000mL)             
            mL                 86.2 g                              ( 1 L )

    = 7.65661 mol hydrocarbon  ?

I have a feeling this lead me nowhere

2C6H14 + 19O2   -->  12CO2 + 14H2O

How many moles of oxygen are needed to burn that amount of hydrocarbon?

[jokerboy111]

Wow I got it...

2C₆H₁₄ + 19O₂    -->  12CO₂ + 14H₂O

  (7.65661 mol hydrocarbon)
(2 mol hydrocarb in equation)    = 3.828305 (ratio)

19 mol O₂ x 3.828305 = 72.7377 mol O₂

Back to the ratio of oxygen to air...

(72.7377 mol O₂)    = .2095     
   (X mol of dry air)     

X mol dry air = 347.197 mol dry air   

plug into PV=nRT and solve for volume....

8490.316517 L dry air

 = 8.49 x 10³ L of dry air 



Thanks so much to all of you who put up with me and contributed..i give you a virtual high five and virtual beer of thanks.