[JDeGra]

Ok, so I'm having trouble getting from liters to mass.  The problem says What volume of oxygen is required to burn completely 5 gallons of a mixture that is 90.0% gasoline (C8H18) and 10.0% ethanol (C2H6O) BY MASS. Well 5 gallons = 18.925 L and I am given a density of gasoline (0.742g/mL) and ethanol (.789g/mL) I have figured out the balanced chemical equations for both reactions but can't really use them until I know how many grams of gas and ethanol I am working with... I know this should be the easy part of the question.  Can anybody give my brain a jumpstart?

[psychoNOT]

So you know 90% of the mixture is gasoline and 10% is ethanol, and the mixture is 18.925L.  This means that 17.0325L of the mixture is gasoline and 1.8925L is ethanol.  Convert these volumes to mL, then multiply that by that volume by the density to get the mass.  I assume you can figure it out from there if you have the equations and can do the stoichiometry.

[Borek]

So you know 90% of the mixture is gasoline and 10% is ethanol, and the mixture is 18.925L.  This means that 17.0325L of the mixture is gasoline and 1.8925L is ethanol.

No - volumes are not additive. Question - as posted - doesn't give enough information.

It can be solved with the assumption you have proposed, but that's only an approximation.

My bet is they expect JDeGra to estimate the density by linear extrapolation. In general that'll be wrong as well.

[JDeGra]

I had already tried finding the mass by calculating 90% and 10% of the volume (17.0325L and 1.8925L respectively) but that didn't bring forth the correct answer.  The only way to find the % of each by mass would be to know the mass of the mixture right? ... Can I do that with the information that I have? Of course I have the temperature and the pressure too. Any other ideas?

[Borek]

The only way to find the % of each by mass would be to know the mass of the mixture right?

Yes.

Can I do that with the information that I have?

No.

Any other ideas?

Have you tried the linear approximation of mixture density?

[JDeGra]

I don't know how to do that. 

[Borek]

100% gasoline - density - 0.742
100% ethanol - density - 0.789

One may expect that 50/50 mixture has a density that is average of these two (that won't be exactly true, but it is better than nothing). Try to calculate what should be density of 90/10 mixture.

[JDeGra]

Oh, I guess I've tried that too.  I've been doing everything short of acrobatics to try to figure this one out.   All the answers I come up with are somewhere between 3.1 x 10^4 L  and   4.3x 10^4 L.  It's a multiple choice question so unfortunately the answers I can choose from are 4.6 x 10^6;  4.7 x 10^6;  1.3 x 10^5; or 1.9 x 10^1.   I think I am going to have to choose "none of the above" for this problem.  Thanks for helping.  At least I know I'm not missing something totally obvious! I hope nobody else in the class got this problem right. LOL

[sjb]

Oh, I guess I've tried that too....

Why not type out what you've got thus far, and we'll see if there are any glaring blunders in your maths?

[Borek]

4.6 x 10^6;  4.7 x 10^6;  1.3 x 10^5; or 1.9 x 10^1

Your results seem to be in the correct ballpark - that is, we don't know exact answer, but it will be something like 4*10⁴ L. None of the answers given is even close, they are all off by orders of magnitude - so in fact it doesn't matter what the exact answer is, none of the above can be correct.

[JDeGra]

None of the answers were correct.  I was right.  The professor said that she only puts "e) none of the above"  as an option to cover her a** when she forgets to make changes and so forth...  This was one of those instances.  Thanks for helping :-)

[billnotgatez]

Did the Prof give you the correct answer that should have been there