Question

The enthalpy change for the reaction
2 H2 + O2 > 2 H20
is -571.6 kJ. Determine the enthalpy change for the decomposition of 24.0g H2O.

My Process
-571.6 is the enthalpy of 2 mols of H2O.
So the enthalpy of 1 mol of H2O will be -285.8.
Since it's decomposition, -285.8 will become +285.8.
Then I would have to convert the 24g into mols. 24/18.02= 1.3 mols
Now I am lost fro here on out

Answer 1

See your CH4 post. It's done the same way but make sure that's for 2 mols H2O.

Answer 2

would I multiply my answer from 1.3(285.8) by 2 then?

Answer 3

To follow the same pattern I followed above it would be
+571.6 kJ for 36 g H2O so how much for 24 g?

(571.6/36) x 24 = ?
You've already divided the 571.6 by 2 so your step is ok as is EXCEPT you rounded and I don't know why? 24/18 = 1.33 mol.

Answer 4

I was trying to keep sig figs. If I were trying to do it my way, I would just multiply 1.33 by 285.8 and that would get me the same answer(almost) at 380.11. Yours is 381.07

Answer 5

If that 24 is correct (and not 24.0) then you're right and I'm wrong and the number of significant figures is 2 so 1.3 is appropriate.
But no matter which way you do the problem you should end up with the same answer if you use the same numbers for the molar masses.