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.

how many moles is 24g?

then multiply that by -571.6kJ

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?

So would it be:

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

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.

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

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.

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

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.

To determine the enthalpy change for the decomposition of 24.0g H2O using the given information, you need to follow these basic guidelines:

1. Write a balanced chemical equation: Start by writing the balanced chemical equation for the decomposition of water (H2O) into hydrogen gas (H2) and oxygen gas (O2).
2H2O(l) → 2H2(g) + O2(g)

2. Determine the molar mass: Calculate the molar mass of water (H2O) by adding the atomic masses of hydrogen (H) and oxygen (O). This will allow you to convert grams to moles.
Molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol

3. Convert grams to moles: Divide the given mass of H2O (24.0g) by the molar mass to convert it to moles.
Moles of H2O = 24.0 g / 18.02 g/mol = 1.33 mol

4. Use the stoichiometry of the balanced equation: Since the balanced equation shows that 2 moles of water produce 2 moles of hydrogen gas, and the given enthalpy change refers to 2 moles of hydrogen gas, we can use stoichiometry to find the enthalpy change for the decomposition of 1.33 moles of water.
Enthalpy change = -571.6 kJ / 2 mol × 1.33 mol = -380.3 kJ

5. Interpret the result: The enthalpy change for the decomposition of 24.0g H2O is -380.3 kJ. The negative value indicates that the reaction is exothermic, meaning it releases energy.

By following these basic guidelines, you can determine the enthalpy change for the decomposition of 24.0g H2O using the given information.