I really wonder why this simple question was not answered for 2 years!

12.7 moles of NH3 "under the assumption that nitrogen is the reactant in excess while hydrogen is the limiting reactant".

Given that the equilibrium constant Kc = 0.061 at 500 K, and:

Kc = n(NH3)^2/[(n0(N2) - n(NH3)/2)*(19-1.5n(NH3))^3];

n0(N2) must be very high to have n(NH3) tends to 19 / 1.5 = 12.666.

3H2---> 2NH3

19 -----> X

X =12.67 mole