How many moles of NH3 can be produced from 19.5mol of H2 and excess N2

This is a simple (as opposed to limiting reagent) stoichiometry problem.

N2 + 3H2 ==> 2NH3
Use the coefficients in the balanced equation to convert mols of anything in the equation to mols of anything else in the equation.
19.5 mol H2 x (2 mol NH3/1 mol N2) = 19.5 x 2/1 = ?

13

39

13

39

13!