Chemistry enthalpy

1. Feb 6, 2014

laurenM

1. The problem statement, all variables and given/known data

Given 2C2H6 $\rightarrow$2C2H4(g) 2H2(G) $\Delta H$=-64.2kJ

$\Delta H$ for the reaction of 6.000g of C2H6(g) to givev C2H4(g) andH2(g) is
2. Relevant equations
$\Delta H$ = $\Sigma products$ - $\Sigma reactants$
q=ct
keeping the units
The enthalpies are not given and are not to be looked up

3. The attempt at a solution
I understand that i have to divide ΔH by 2 and multiply by 6g
but that gives me a unit of kJ*g so i need to divide by grams to get the units right... my solution book tells me to divide by 30g but i am unaware as to where that comes from

2. Feb 6, 2014

BOYLANATOR

You are given the enthalpy for a reaction of one mole. How much does a mole weigh in grams of this material? That would be my thinking, but the book has a slightly different number than I would expect

3. Feb 6, 2014

Staff: Mentor

Book is right.

Enthalpy is given for two moles, not for one. But generally speaking you are on the right track, it is abut finding number of moles.

4. Feb 6, 2014

laurenM

The answer from the book is-6.41 kJ. I found the mols of C2H6 and used the ratio to figure out the sum of mass of reaction and products to get to get 12.00g this gives me the answer of -16.05 kJ.
=-64.2/2 to get it to a 1 to 1 ratio
=-32.1*6 multiply by 6.00g
=-192.6/12
=-16.05 kJ
if you divide by 30 instead of 12 you get the right answer i just don't get where it comes from

5. Feb 6, 2014

BOYLANATOR

I don't follow your logic here. You worked out the enthalpy would be -32.1 kJ for a one mole reaction. That's ok.

But in this case it's not one mole of C2H6, it's 6 grams. What fraction of a mole is 6 grams?

6. Feb 6, 2014

Staff: Mentor

Just make sure that the units cancel properly:

$$(32.1 \frac{kJ}{mole})(\frac{1mole}{30grams})(6grams)=? kJ$$