Answer:
![4molNH_3](https://img.qammunity.org/2022/formulas/chemistry/high-school/5opbt6623dyt0lzav58ytt5hx1etsmyet1.png)
Step-by-step explanation:
Hello there!
In this case, according to the given information it will be firstly necessary to set up the chemical equation taking place:
![N_2+3H_3\rightarrow 2NH_3](https://img.qammunity.org/2022/formulas/chemistry/high-school/pjb2nctitg3minc8rtw5p32uw2v35eq4eg.png)
We infer we need to calculate the moles of NH3 by using both of the moles of N2 and H2 at the beginning, in order to identify the limiting reactant:
![n_(NH_3)=6molN_2*(2molNH_3)/(1molN_2)=12molNH_3\\\\ n_(NH_3)=6molH_2*(2molNH_3)/(3molH_2)=4molNH_3\\](https://img.qammunity.org/2022/formulas/chemistry/high-school/qapzcitwlszkch3hth3s9et4iyq8sm1ozg.png)
Thus, since hydrogen yields the fewest moles of ammonia, we conclude that we are just able to yield 4 moles of NH3.
Regards!