[Mitch]

The delta H is equal to the heat of neutralization divided by moles of the limiting reagent.

[meme]

okay... lemme just make sure I got this all straight:

• Heat evolved:

.05 mol NaOH x (38.99g/1mol) = 1.95g NaOH
heat evolved: (4.18)(1.95g)(6.9C) = 56.2J

• Amount of OH^- reacted:
  so... since it's the same as the amount of NaOH - its .05 mol OH^- right?
  
• Amount of H₂O formed:
  1:1 mole ratio - therefore .05 mol water
  
• Heat evolved per mole of H2O, (heat of neutralization):

56.2 J/ .05 mol = 1124.5 J/mol ?? [/list]

*squeezes eyes shut, crosses fingers, and hopes all that's right*

[Mitch]

Nope you were close though.

You have the right answer but the .05moles at the last step isn't coming from the moles of water but is the moles of hydroxide. It is the limiting reagent that matters when you divide not the water.

But, yeah it all looks right to me.

[meme]

56.2 J/ .05 mol HYDROXIDE = 1124.5 J/mol

so everything is right now, right?  

anyway, thanks a million! It would have taken me forever without your help.  

[Mitch]

looks good.

[meme]

Is there any way I can donate to the site?

[Mitch]

We might start selling merchandise in the coming month, like little periodic tables and the like. But we don't accept straight donations. Although, the sooner Greg makes the logo, the sooner we can start selling some stuff.