Answer:
[tex]29.6kgNH_3[/tex]
Explanation:
Hello,
In this case, it is firstly necessary to know the limiting reactant via the shown below procedure:
[tex]n_{N_2}^{available}=31.5kgN_2\frac{1kmolN_2}{28kgN_2}=1.125kmolN_2[/tex]
[tex]n_{N_2}^{consumed\ by\ H_2}=5.22kgH_2*\frac{1kmolH_2}{2kgH_2}*\frac{1kmolN_2}{3kmolH_2}=0.87kmolN_2[/tex]
In such a way, the hydrogen is the limiting reagent as 5.22 kg of it do not consume all the available nitrogen, therefore, the theoretical yield of ammonia turns out as shown below:
[tex]m_{NH_3}=0.87kmolN_2*\frac{2kmolNH_3}{1kmolN_2}*\frac{17kgNH_3}{1kmolNH_3}=29.6kgNH_3[/tex]
Best regards.