Answer: 2.917 moles of [tex]N_2[/tex] are present in excess.
Explanation: For the given reaction:
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
Number of moles of [tex]N_2[/tex] = 3.5 moles
Number of moles of [tex]H_2[/tex] = 1.75 moles
Here, the [tex]H_2[/tex] is considered as limiting reagent as it limits the amount of product formed.
By stoichiometry, 3 mole of [tex]H_2[/tex] require 1 mole of [tex]N_2[/tex] to produce 2 moles of [tex]NH_3[/tex].
So, 1.75 moles of [tex]H_2[/tex] will require = [tex]\frac{1}{3}\times 1.75\text{ moles of }N_2[/tex]
Moles of [tex]N_2[/tex] reacted with [tex]H_2[/tex] = 0.583 moles
[tex]\text{Moles of }N_2\text{ in excess}=(3.5-0.583)moles[/tex]
= 2.917 moles
