[Kipper]

Hello,

This is actually a 3 part question.  It is the last part that is giving me trouble.

Question C asks the following:

A 10.0 gram sample that is 30% C2H6 and 70% C4H10, by mass, is burned in excess oxygen.  How much energy is released?
The answer for this is actually 4.6 x 10^2 kj but I can't figure out how they got this.

I'm not sure if the answers from (a) and (b) are relavant but I'll post them just in case:

a) calculate the enthalpy change for the complete combustion of
C4H10: answer is -2657 kj/mol

b) calculate the enthalpy change for the complete combution of C2H6:
answer -1428.4 kj/mol

If someone could enlighten me I would greatly appreciate it.
Thank you.
Kasey

[ARGOS++]

Dear Kipper;

Yes!, - The answers a) and b.) are very relevant to get the heat released.
Use the percentage(w/w) and show us your try, so we can tell you where you may go wrong.

Good Luck!
                    ARGOS⁺⁺

[Kipper]

Well at first I confirmed that C2H6 was 3 grams and C4H10 was 7 grams and then I found the molar mass of each and calculated the moles per molecule.

For C2H6 I got 0.0997 moles and then for C4H10 I got 0.1204 moles.  That's where I got stuck.  I was thinking that because everything had to be in moles (usually) it was important for me to calculate them.

I then tried to calculate delta H for each one individual and then was thinking I could add the delta H's together but it didn't work out.  Here's what I did:

C2H6:
delta H = -1428.4 x 0.0997 moles and then divided by current moles of 2
          = -71.21 kj

C4H10
delta H = -2657 x 0.1204 and then divide by a mole of 1
          = -319.90 kj

Adding them together gives -391 kj.

Am I completely off base in how I'm trying to solve this?

[ARGOS++]

Dear Kipper;

Please can you tell me why you divide?  I don’t see any reason for.

Good Luck!
                    ARGOS⁺⁺

[Kipper]

Funny I'm not sure why.  I think I was trying to get the total answer into 1 mole, so I divided by the mole of 2 (which is based on the equation I had - equation showed 2 moles of C2H6).

However, if I don't divide at all and add the two amounts together I get the right answer of -4.6 x 10^2.

Yahoo!!!

Honestly I have no idea how I came up with the right answer but I did.

[ARGOS++]

Dear Kipper;

For this part there is no equation required at all, because you can use answers a.) and b.) instead.

So finally you did correct.

Good Luck!
                    ARGOS⁺⁺