Question

I need a recap of how to do the question below. I just need the basic guidelines:

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

Answer 1

how many moles is 24g?

then multiply that by -571.6kJ

Answer 2

If I was given a question like this, I would just find the moles of the mass given and multiply it by the enthalpy change?

Answer 3

So would it be:

24/18=1.33
1.33(-571.6)=-761 kJ

Answer 4

I believe the previous answer is based on that being -571.6 kJ/mol but actually it is -571.6 kJ/2 mol or -285.8 kJ/mol (and it says -571.6 kJ/reaction and that's for 2 mols). Then multiply by 1.33.

Answer 5

What I am getting at is that the delta H given is for the product of the equation. Now, considering the number in front of it, you should divide the delta H by that number to get the delta H for one mol. From there you would multiply it by the mol of the product.
So the answer would be -380 then

Answer 6

I think you are right; however, I don't know that I will go along with the 380 unless that's just a quickie number. The 571.6 has four places in it and if the other numbers have that many you are allowed more than 3 places in the answer. If I leave all of the numbers in the calculator I get more than 380.

Answer 7

I did round to fit the sig figs but wasn't sure if I was to add decimals or not

Answer 8

You're right. 24.0 is the limiting number so you are allowed only three s.f. Probably I would report -381 kJ since -571.6 x 1/2 x (24.0/18.0) gives -381.066 kJ which rounds to -381 kJ to 3 s.f.