If I have 6 mol of N2 and 12 mol of H2 to yeild ammonia, which chemical is in excess and by how much?
N2 + 3H2 ==> 2NH3
Use dimensional analysis and the coefficients.
for N2.
6 mol N2 x (2 moles NH3/1 mole N2) = 12 mol NH3 produced.
For H2.
12 mol H2 x (2 moles NH3/3 moles H2) = 8 moles NH3 produced.
Therefore, hydrogen is the limiting reagent and N2 is in excess because in limiting reagent problems the SMALLER value of the product formed is ALWAYS the correct answer. The reason is that you can't produce more than the limiting reagent allows.
How much N2 is in excess. Now that you know the H2 is the limiting reagent, use the same kind of factor to determine how much N2 is needed to react with the all of the H2.
12 moles H2 x (1 moles N2/3 moles H2) = 4 moles N2 needed. You had 6 initially; therefore, there must be ?? that doesn't react.