Two equations and two unknowns.
Let X = g CH4
and Y = g C2H6
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eqn 1 is X + Y = 6.79g
eqn 2 is heat generated by CH4 + heat generated by C2H6 = 369 kJ.
heat from CH4 is 890.3 kJ/mol x (X/16)
heat from C2H6 is 1569.7 kJ/mol x (Y/2*30) so all of that together is
X+Y = 6.79
(890.3)(X/16) + (1569.7)(Y/2*30) = 369
Solve for X and Y, then
%CH4 = (mass CH4/6.79)*100 = ?
%C2H6 = (mass C2H6/6.79)*100 = ?
When a 6.79-g mixture of methane, CH4, and ethane, C2H6, is burned in oxygen at constant pressure, 369 kJ of heat is liberated. What is the percentage by mass of CH4 in the mixture? The standard enthalpies of combustion for CH4 and C2H6 are -890.3 kJ mol-1 and -1569.7 kJ mol-1, respectively.
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