The product mixture would contain no $H_2$ molecules, 3 $N_2$ molecules, and 4 $NH_3$ molecules.
Work Step by Step
The reaction is as follows: $$N_2+3H_2\Rightarrow 2NH_3$$ Therefore, $H_2$ is the limiting reactant and all $H_2$ is consumed. Furthermore, for every 3 $H_2$ molecules consumer, one $N_2$ molecule is consumed. Therefore, because 6 $H_2$ molecules are consumed, 2 $N_2$ molecules are consumed with 4 $N_2$ molecules remaining. Similarly, because for every 3 $H_2$ molecules consumed, two $NH_3$ are produced. Therefore, because 6 $H_2$ molecules are consumed, 4 $NH_3$ molecules are formed.