# Multiply by the number of moles of that substance (1 Viewer)

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#### 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

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

P

#### PPonte

##### Guest
How do I figure out how many moles to mutiply by?
Is this an exercise?
If it is, they must say the number of moles of H2O and the number of moles of NH3.

And what does the L or G mean for the equation?
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.

#### 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

P

#### PPonte

##### Guest
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.

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#### 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.

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