Answer :
Answer:
Follows are the solution to these question:
Explanation:
Given equation:
[tex]N_2+3H_2 \longrightarrow 2NH_3[/tex]
In this equation:
[tex]1 N_2[/tex] gives [tex]= 2NH_3[/tex]
so,
[tex]3N_2[/tex] gives= [tex]2 \times 3 = 6 NH_3[/tex]
similarly:
[tex]3H_2[/tex] gives [tex]= 2NH_3[/tex]
So, [tex]6H_2[/tex] gives = [tex]\frac{2}{3}\times 6=4NH_3[/tex]
Its limited reagent is =[tex]N_2[/tex]
The amount of [tex]NH_3[/tex] molecules were formed = 4.
and the amount of [tex]H_2[/tex] excess molecules are= 1