Question

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?

PLEASE HELP! I do not know where to even start!

Answer 1

This is a long one and will take your complete concentration to follow but here is what you do. Here are the combustion equations.
rxn 1. CH4 + 2O2 ==> CO2 + 2H2O
rxn 2. C2H4 + 3O2 ==> 2CO2 + 2H2O
Next you calculate the heat evolved with rxn 1 and rxn 2. These are done by looking up the dHformation in tables in your text/notes. I will call these dH1 and dH2.
dH1 = (n*dHf products) - (n*dHf reactants)
dH2 = (n*dHf products) - (n*dHf reactants)
You will get a number for dH1 and dH2.
You use this information to set up and solve two equations simultaneously. You will need the two equations which I will call eqn 1 and eqn 2.

Let X = mass CH4
and Y = mass C2H4
---------------------
eqn 1 is X + Y = 10 grams.

eqn 2 is made up of the heat generated by rxn 1 + heat generated by rxn 2 to produce 520 kJ. I will let mm stand for molar mass. The heat generated by CH4 will be (X/mm CH4)*dH1.
The heat generated by C2H4 will be
(2Y/mm C2H4)*dH2 and you put these together to make eqn 2 (which is to be solved with eqn 1).
(X/mm CH4)*dH1 + (2Y/mm C2H4)*dH2 = 520

Solve eqn 1 and eqn 2 for X and Y then plug into the % formula to find % CH4 and %C2H4

%CH4 = (X/10)*100 = ?
%C2H4 = (Y/10)*100 = ?

Answer 2

What does n equal in the dH equations? number of moles?