What is the Percentage by Mass of CH4 in the Mixture?

[ElementUser]

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: CH_4(g) +2 O_2(g) -> CO_2(g) + 2 H₂O_(l) \DeltaH= -890.4 KJ

C₂H_6(g) + \frac{7}{2} O_2(g) -> 2 CO_2(g) + 3 H₂O_(l) \DeltaH= -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.

\frac{0.57}{1.57}=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~

 

[Borek]

17.7% doesn't look as a correct answer to me.

 

[ElementUser]

Hm ok, I was just wondering if there was another method to solve this or if the question's answer was just wrong.

Thanks for your confirmation :)

 

[Borek]

There is other method.

You have two unknowns - mass of ethane and mass of methane. Their sum is 10.00. Write other equation for amount of heat produced when you burn the mixture. These are two equations in two unknowns.