Answer:
121.6 of [tex]NH_{3}[/tex] ,Nitrogen is limiting and Hydrogen is in excess.
Step-by-step explanation:
The reaction involved in this question is known as Haeber's Process and the reaction is as follows
[tex]N_{2} + 3H_{2} -->2 NH_{3}[/tex]
We need to find the moles of each to determine the total product, the excess reagent and limiting reagent.
No. of moles of[tex]N_{2}[/tex]=[tex]\frac{100}{28}[/tex]=3.57 mol
No. of moles of [tex]H_{2}[/tex]=[tex]\frac{100}{2}[/tex]=50 moles
We need moles of [tex]H_{2}[/tex] and [tex]N_{2}[/tex] in the ratio of 3:1.
Therefore, we can clearly see that [tex]H_{2}[/tex] is excess. And [tex]N_{2}[/tex] is the limitng reagent.
As, [tex]N_{2}[/tex] is the limiting reagent a total of 2 * 3.57 moles of [tex]NH_{3}[/tex] will be formed.
Therefore, total mass of [tex]NH_{3}[/tex] =7.14*17=121.6 g