[crazypakithug]

NEEDS TO BE SOLVED USING STIOCHIOMETRY
When a 10.0g sample of a mixture of CH4 and C2H6 is burned in excess oxygen, exactly 525KJ of heat is produced. What is the percentage by mass of CH4 in the original mixture.
Heat of combustion of CH4 = -890.4KJ/mol
Heat combustion of C2H6 = -1560.0KJ/mol
Hi this is my first post so I hope i did everything correctly I would appreciate it if anyone can help me do this problem.
I got the equations:
     CH4 + 2O2 --> CO2 + 2H20
     2C2H6 + 7O2 --> 4CO2 + 6H2O
then i think we combine them to get:
     CH4 + 2C2H6 + 9O2 --> 5CO2 + 8H2O
but after this i dont get want to do...any help would be appreciated. I know this goes against the rules but I got this problem today and my teacher wants it tomorrow, so I would appreciate it if anyone can help me do it today.
thanxs for your help

[Sev]

I solved it using simultaneous eqns.

eg.
890.4x + 1560y = 525

using molar masses, 16.0425x+30.07y=10

(x and y are moles of methane and ethane respectively)

[crazypakithug]

can u explain wat simultaneous eqations are...how would you solve it using stiochiometry

[Sev]

simultaneous eqns are a set of eqns that can be solved using the other eqns of the set.
http://en.wikipedia.org/wiki/Simultaneous_equations

...how would you solve it using stiochiometry

hmmm...I can't think of any other method to solve it.  You need to use a bit of stoichiometry to solve it using simultaneous eqns.

[crazypakithug]

thx for your help, would X =1001/3632 or .103 and y=.1017

[Sev]

thx for your help, would X =1001/3632 and y=.1017

I got x (mole of methane) = 0.1068.  You don't need to work out y.

Using this value of x, the mass of methane in the sample will be 1.714g.  So the composition is 17.14%.

[crazypakithug]

thx for your help

[crazypakithug]

does anyone know how to do this problem using stiochiometry

[Borek]

IMHO simultaneous equations are the only possible approach. One for mass and one for heat. Stoichiometry - while present in the background - is not giving enough information.

[crazypakithug]

thanxs for all your help everyone