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Homework Statement
When a 10.0g sample of a mixture of CH4 and C2H6 is burned in excess oxygen, exactly 525 KJ of heat is produced. What is the percentage by mass of CH4 in the original mixture?
Given: CH4(g) +2 O2(g) -> CO2(g) + 2 H2O(l) [tex]\Delta[/tex]H= -890.4 KJ
C2H6(g) + [tex]\frac{7}{2}[/tex] O2(g) -> 2 CO2(g) + 3 H2O(l) [tex]\Delta[/tex]H= -1560.0 KJ
Homework Equations
N/A
The Attempt at a Solution
I take the ratio of the enthalpy of CH4 to C2H6. To do this, I do -890.4/-1560.0. This gives me approx. 0.57 (note that I keep as many digits as possible when doing the actual calculations). Therefore, for every one mol of C2H6 burned, 0.57 mol of CH4 is burned.
I then find out how much of the % of the mixture is burned for the CH4.
[tex]\frac{0.57}{1.57}[/tex]=36.3%
I proceed to multiply the 525 KJ by this percentage and then convert to grams of CH4 to get 3.44 g of CH4. Finally, I find the % mass by dividing my 3.44 g of CH4 by 10.0g of the original mixture, multiply it by 100% to obtain 34.4% of the original mixture as my final solution.
However, the answer appears to be 17.7%; precisely half of the % mass I acquired. Can someone help point out on what I did wrong?
Any help is appreciated. Thanks in advance~