# Multiply by the number of moles of that substance

1. Mar 21, 2006

### sugarandspice

How do I figure out how many moles to mutiply by?

It says "multiply by the number of moles of that substance"

How do I figure that out?

Here's where I am:

H2O(l)= - 285.830 kJ/mol x ___ mol

NH3(g)= -46.11 kj/mol x ___ mol

---

And what does the L or G mean for the equation?:grumpy:

2. Mar 21, 2006

### PPonte

Is this an exercise?
If it is, they must say the number of moles of H2O and the number of moles of NH3.

l and g are the states of water and ammonia and they stand for liquid and gaseous, respectively. In this situation, I think they are not important, but I don't understand well this exercise.

3. Mar 21, 2006

### sugarandspice

This is the question:

Balance the equation and Calculate the enthalpy change for the following reaction:

NH3(g) + O2(g) --> N2(g) + H2O(l)

NH3(g)= -46.11 kj/mol

O2(g) = 0 kJ/mol

N2(g) = 0 kJ/mol

H20(l)= - 285.830 kJ/mol

I know that to begin, I need to multiply the numbers above by the amount of moles in the substances. I think. LOL

4. Mar 21, 2006

### PPonte

First, we balance the equation:

2NH3(g) + 1.5O2(g) --> N2(g) + 3H2O(l)

Enthalpy change = enthalpy of products - enthalpy of reactants
Enthalpy change = 2 mol x -46.11 kj/mol - 3 mol x - 285.830 kJ/mol
Enthalpy change = -92.2 + 857.49
Enthalpy change = 765.29 kJ

Note: I haven't learnt yet to resolve enthalpy change problems in school like this, but is the only way I see to resolve it. I seriously advise you to wait for a confirmation or correction of this resolution. I just tried to help.

Last edited by a moderator: Mar 21, 2006
5. Mar 21, 2006

### sugarandspice

Looking over my notes, it looks like you hit it right on! I'm still a *little* fuzzy but I can't thank you enough for helping me out. I was gonna have to submit this with no idea what I was doing.. :) Any help at all is better than no help. Thankyou SO much for your help.