[rijo664]

OK i need some of your help. the questions says is

6.0 moles of N2 thats (N and the subscript 2) are mixed with 12.0 moles of H2 according to the following equation:

The equation is already balanced:

N2(g)+ 3H2(g) yields 2NH3 (g)

The questions asks for what is the chemical in excess? is it H2 or N2
I solved the problem but i want to make sure if it is right. this is what i did

6.0 moles of N2  (2 mol NH3 / 1 Mol of N2)= 12 mol of NH3
12.0 moles of H2 (2 mol NH3 / 3 mol H2)= 8 mol of NH3

I believe the answer is H2 because given amount is 12.0 moles and i got 8 mol of NH3 is it right or wrong please tell me.
 

[rijo664]

can anybody tell me what i put up there is right or wrong, or is that a violation of the chemistry forums rule i showed you my work. so could you please tell if it is right or not.

[Sev]

N₂ is in excess.  You have 12 mole of H₂ - this only reacts with 12/3 = 4 mole N₂. 

[Borek]

Do you really think your questions will be answered in 7 minutes?

Question is OK and is not against rules, but you will need some patience, sometimes even a day won't be enough.

Question asks about EXCESS reagent, so while your calculations look OK, your answer looks wrong.

[rijo664]

thanks i don't expect anyone to reply back within 7 minutes but if u get 11 views on that post within 5 minutes then people should be able to put a reply on there.

[rijo664]

Let me ask u another question

Theoretically, how many moles of NH3 will be produced

I believe it's 2 Moles of NH3 because that's how many there are in the balanced equation. Is it right if so

How should i get the actual yield of NH3 if the percent yield is 80%, how many moles of NH3 are actually produced. I am kind of stuck here if the theoretical is 2 then how will i find the actual

I know the formula for percent yield is

actual yield/theoretical yield X 100

help would be appreciated.

[Sev]

H₂ is the limiting reagent (so all 12 mole of H₂ will react).  From rxn eqn, 3 mol H₂ reacts to give 2 mol NH₃; so 12 mole will give 8 mole NH₃.

If yield is 80%, 8*0.8 = 6.4mole of NH₃ forms.

[agrobert]

This is a limiting reagent problem.  Ask yourself which reagent H₂ or N₂ is in excess, then the other will be limited.  Your calculations are correct but your interpretation is wrong.  BTW subscript is a button option when posting.