I have obtained the correct answer using all but one equation, the enthalpy of solution, which I don't know how to use at all. This is how I arranged my equations to give the correct answer:

1. 1/2 O

_{2(g)} O

_{(g) } ΔH = 125.5 kJ

2. 1/2 H

_{2(g)} H

_{(g)} ΔH = 217.5 kJ

3. Na

_{(s)} Na

_{(g) }ΔH = 109 kJ

4. O

_{(g)} + H

_{(g)} O--H ΔH = -465 kJ

5. Na

_{(g)} + O

_{(g)} Na--O ΔH = -255 kJ

6. NaOH

_{(s)} Na

_{(s)} + 1/2 O

_{2(g)} + 1/2 H

_{2(g) }ΔH = 427 kJ

I don't know how to show the slant symbol for cancelling terms but if you add all equations together on paper, the final equation should be

NaOH(s) + O(g) O--H + Na--O. This is where I get confused, how do you remove the

O(g) term on the reactant side so that the net reaction is

NaOH(s) O--H + Na--O OR

NaOH(s) NaOH(g)