I am sure this will be an easy fix for you guys. I understand how to find Enthalpy, but the part I do not understand is how the 3 moles of NH3 plays into it or how to convert/figure it into the process. I know to ignore excess O2.

Determine the standard change in enthalpy for the following reaction:

4 NH3 (g) + 7 O2 (g) → 4 NO2 (g) + 6 H2O (l) when 3 moles of NH3 is reacted in excess oxygen.

Question

I am sure this will be an easy fix for you guys. I understand how to find Enthalpy, but the part I do not understand is how the 3 moles of NH3 plays into it or how to convert/figure it into the process. I know to ignore excess O2.

Determine the standard change in enthalpy for the following reaction: 

4 NH3 (g) + 7 O2 (g) → 4 NO2 (g) + 6 H2O (l) when 3 moles of NH3 is reacted in excess oxygen.

dtan5457 · Answer

You need to goto your reference table (W-4) that gives the enthalpy formation heats per each element, compound. 

Add them all up accordingly so let's say NO2 is 33.2 ( it is), multiply that by 4.


Then subtract as

products (total)- reactants

anonymous · Answer

Ah, thanks for the insight! However, the part I am confused about is how the equation changes because of the addition " when 3 moles of NH3 is reacted."

dtan5457 · Answer

@abb0t @Callisto

anonymous · Answer

That is simple, @Laylaj. They just want you to work with 3 mole and all you have to do is determine the moles of the other substances using proportion.
You can see that the ratio of NH3 to Oxygen is 4:7, so if you have 3 of NH3, you shall have 3*7/4=21/4 mol of Oxygen .
The ratio of NH3 to NO2 is 1:1, hence given that you have 3 of NH3, you shall also have 3 of NO2.
The ratio of NH3 to H2O is 1:1.5, so if you have 3 of NH3, you shall have 4.5 of H2O.
Now these will be your new moles and work with them.