A 10.0 gram sample of a mixture of CH4 and C2H4 reacts with oxygen at 25°C and 1 atm to product carbon dioxide gas and liquid water. If the reaction produces 520 kJ of heat, what is the mass percentage of CH4 in the mixture?
I am all out of tries on the homework, but I would like to know the answer so I can study for my exam!
PLEASE HELP ME, NO ONE ELSE CAN!
5 answers
I thought I worked this for you a couple of days ago.
I tried to use how you answered for someone else but I got it wrong and used all my tries so I would like the real answer to see where I went wrong!
Thanks for your consideration!
Thanks for your consideration!
I never worked it through the other day but in setting it up today I think I may have erred in that first post. Give me some time and I'll try to come up with an answer. You can help if you will post what you calculated for dH1, dH2 and the equations.
dH1 i got -890 kJ/mol
dH2 i got -1411 kJ/mol
dH2 i got -1411 kJ/mol
I came up with 3.11 g CH4 and 6.89 g C2H4.
I checked those answers this way.
The dHrxn for CH4 I found as -890.3 kJ/mol.
and dHrxn for C2H4 as -1410.09 kJ/mol.
Heat produced by 3.11g CH4 is 890.3 kJ/mol x (3.11/16) = 173 kJ.
Heat produced by C2H4 is 1410.09 x (6.89/28) = 347 kJ.
Then heat from CH4 + heat from C2H4 = 173 kJ + 347 kJ = 520 kJ which is what we had in the problem so I feel certain those answers above are correct.
I checked those answers this way.
The dHrxn for CH4 I found as -890.3 kJ/mol.
and dHrxn for C2H4 as -1410.09 kJ/mol.
Heat produced by 3.11g CH4 is 890.3 kJ/mol x (3.11/16) = 173 kJ.
Heat produced by C2H4 is 1410.09 x (6.89/28) = 347 kJ.
Then heat from CH4 + heat from C2H4 = 173 kJ + 347 kJ = 520 kJ which is what we had in the problem so I feel certain those answers above are correct.
2 answers