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**1. Homework Statement**

When a 10.0g sample of a mixture of CH

_{4}and C

_{2}H

_{6}is burned in excess oxygen, exactly 525 KJ of heat is produced. What is the percentage by mass of CH

_{4}in the original mixture?

Given: CH

_{4(g)}+2 O

_{2(g)}-> CO

_{2(g)}+ 2 H

_{2}O

_{(l)}[tex]\Delta[/tex]H= -890.4 KJ

C

_{2}H

_{6(g)}+ [tex]\frac{7}{2}[/tex] O

_{2(g)}-> 2 CO

_{2(g)}+ 3 H

_{2}O

_{(l)}[tex]\Delta[/tex]H= -1560.0 KJ

**2. Homework Equations**

N/A

**3. The Attempt at a Solution**

I take the ratio of the enthalpy of CH

_{4}to C

_{2}H

_{6}. 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 C

_{2}H

_{6}burned, 0.57 mol of CH4 is burned.

I then find out how much of the % of the mixture is burned for the CH

_{4}.

[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 CH

_{4}to get 3.44 g of CH

_{4}. Finally, I find the % mass by dividing my 3.44 g of CH

_{4}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~