Chemistry help...

[xjohnieboyx]

Somebody help me out if you know anything about chemistry:

3 Liters of C2H6 are mixed with 3.5 Liters of O2 (Oxygen). As a result, CO2 (gas) and H2O (gas) are formed. Given, all of the volumes are measured under the same conditions (temperature, pressure). What is the overall volume of the two gases at the end of the equation?

Equation looks like this:

C2H6 + O2 ----> CO2 + H2O

I'm guessing the first step is to balance the equation, but I'm having trouble finding the correct coefficients...



PS: Hi guys i haven't been here in a while lol

[Fantasticone]

2C2H6 + 7O2 ----> 4CO2 + 6H2O

Hoping that my chem knowledge still works.

[xjohnieboyx]

Actually, Fantasticone, that's what I came up with as well.

[PinoySkillz]

Hmmm...it's been a while, but this looks like a Gas Law equation:

P1*V1/T1 = P2*V2/T2

But if that is the right equation then the variables for P2 and T2 must have been given to you to solve for V2. Did they give you the variables for those 2? If not then I'm probably wrong...

[GuidoHunter]

(Liters of C2H6) * (m^3 C2H6/Liters C2H6) * (density C2H6) * (1/Molar mass C2H6) * (4 moles CO2 / 2 moles C2H6) * (Molar mass CO2) * (1/density CO2) = Volume CO2 (in m^3)

Then do the same thing (calculate the volume of CO2), but with the values for Oxygen instead of C2H6.

Then take the equation that gets the SMALLER result of CO2 (because that reactant will be used up first), and plug in values of H2O instead of CO2, thus finding the final volume of H2O.

Add the two volumes together, and you have your answer (be sure to convert back to liters if you have to).

EDIT: for the equation at the top, all you have to do is cancel units until you have what you want:
[L]*[m^3/L]*[kg/m^3]*[mol C2H6/kg]*[mol CO2/mol C2H6]*[kg CO2/mol CO2]*[m^3/kg] = [m^3]
Notice how everything cancels except for the volume of CO2. This is how simple stoichiometry works.

--Guido

http://andy.mikee385.com

[xjohnieboyx]

cheers Guido