N2(g) + 3H2(g) -> 2NH3(g)

My friend needs help on this. We've tried everything and everyone she asks doesn't know. Can someone help?

6.0 mol of N2 are mixed with 12.0 mol of H2 according to the following equation:
N2(g) + 3H2(g) -> 2NH3(g)

Which chemical is in excess? What is the excess in moles?

Answers appreciated. :)

bal. rxn.: N2(g) + 3 H2(g) -> 2 NH3(g)
1 mole N2 reacts with 3 moles of H2 (1:3)

6 moles N2 and 12 moles H2 (1:2)
N2 is in excess
12 moles H2/3= 4 moles N2 required
2 moles N2 is in excess

From the balanced equation, 6.0 mol of N2 produces 2(6.0) = 12.0 mol of NH3, while 12.0 mol of H2 produces (2/3)(12.0) = 8.0 mol of NH3. Since H2 produces less NH3 than N2 does, N2 is the chemical in excess.

I hope this helps!