How many moles of #N_2# reacted if 0.75 mol of #NH_3# is produced?

1 Answer
Nov 3, 2016

Answer:

#1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)#; #0.38*mol# #N_2# reacted.

Explanation:

Given the stoichiometrically balanced equation, if #0.75*mol# ammonia were produced, then at least #(0.75*mol)/2# dinitrogen reacted. Is this clear?