[james bond 3]

How many L of oxygen would be required to burn 1.00g of Octane?

When:
     Pure oxygen is 740mm Mercury and 24º C
     Octane is 8 carbons and 18 hydrogens.

It would be really helpful if you could show how to figure the problem out.
Thanks!

[ARGOS++]

Dear James bond 3,

That’s a normal “Stoichiometry Problem”.

So you may apply the recipe/scheme from the Topic “Stoichiometry Problem”:   "Stoichiometry Problem”
Finally you translate the mole amount of Oxygen by:  "Ideal Gas Law” into the correct Volume!

I hope it may be of help to you.

Good Luck!
                    ARGOS⁺⁺

[james bond 3]

thank you so much this is a great begining but I am a bit stuck on the set up:
 
8C + 18H ----> ?

[ARGOS++]

Dear James bond 3,

Do it easier! ,  ─  Because as you know the formula for Octane is C₈H₁₈.

Can you now complete the Line A.) ?

 A.)          C₈H₁₈    +      O₂   ------>    ? ?   +  ? ?

Good Luck!
                    ARGOS⁺⁺

[phmonkey]

just to get things rolling along
A.)          C8H18    +      O2   ------>    ? ?   +  ? ?
              2C8H18  +    25 O2 ----->   18H2O  +  16CO2
    C=16, H=36, O=50                                      H=36, C=16, O=32+18=50

[ARGOS++]

Dear Phmonkey,

Perfect!,   ─   Even when James bond 3 was asked.

So we have now:
A.)           C₈H₁₈    +         O₂   ------>        CO₂       +        H₂O
B.)        2 C₈H₁₈    +    25 O₂   ------>   16  CO₂       +   18 H₂O

James bond 3,  can you now give the values for the line C₁.)?

Good Luck!
                    ARGOS⁺⁺

[james bond 3]

okay, so just making sure- you would get co2 and h20 because your burning a hydrocarbon? and then you balance... so


2C8H18  +    25 O2 ----->   18H2O  +  16CO2



and now im a little lost. so how do i  move on from here?

[james bond 3]

i dont believe i have a known mass. only OXYGEN AT 740 ml HG which is a pressure i believe.

[phmonkey]

u have 740 mmHg, its 760mmHg = 1atm, and stp is 1atm 0 degrees C, hopes this helps

[ARGOS++]

Dear James bond 3,

Great!,   ─   You did also perfect!

Not too fast!:
For C₈H₁₈ as an Example you have to summarise 8 times the Atom mass of Carbon plus 18 times the Atom mass of Hydrogen.

How much is that?  And for the other reaction partners/products?

Good Luck!
                    ARGOS⁺⁺

[james bond 3]

C8H18 = 114 MOLES

O2= 36 MOLES

H20 = 18 MOLES

CO2 = 48 MOLES

but what baout the coefficicents in the front of them? how do they get factored in?

[ARGOS++]

Dear James bond 3,

Nearly Perfect!,  ─   Only g / mole, that’s enough!

So we have now:
A.)           C₈H₁₈    +         O₂   ------>        CO₂       +        H₂O
B.)        2 C₈H₁₈    +    25 O₂   ------>   16  CO₂       +   18 H₂O
C₁.)          114 g/m           32 g/m                44 g/m            18 g/m       
C_m.)          1.00 g
C₂.)            ?  m
D.)             ?!   

I have already prepared the Line C_m.), because the only thing you know is the 1.00 g for Octane.
You know what you have to do for the only value in Line C₂.)?: Convert the 1.00 g into the corresponding “Number” of Moles.             

And if we have done also the Line D.), we are two steps before the End of the Stoichiometry!
So please keep patience.                                                                                     


Good Luck!
                    ARGOS⁺⁺

[james bond 3]

Would this be the answer to the question?:

1.00g C8H18 (1molC8H18/114gC8H18) (25molO2/2molC8H18)= .110 mol O2
 
PV=nRT
 
740mmHg(1atm/760mmHg)= .974 atm
273+24= 297K
 
(.974atm) (V) = .110mol (.08206 L atm/K mol) (297K)
 
V=2.75L

[ARGOS++]

Dear James bond 3,

Yes!,   ─    That looks correct, because after the two last Steps you would have got the result of 0.1096 moles of Oxygen on Line C₂.).
In the Scheme you would be able to verify your result for the moles of Oxygen.

The other calculation looks ok to me!

Good Luck!
                    ARGOS⁺⁺

[james bond 3]

thank you so much!