For the reaction represented by the equation #N_2 + 3H_2 -> 2NH_3#, how many moles of nitrogen are required to produce 18 mol of ammonia?

2 Answers
Jan 26, 2016

Answer:

#9molN_2#

Explanation:

The reaction is:

#N_2(g) + 3H_2(g)->2HN_3(g)#

The molar ratio between Nitrogen and ammonia is #1:3#, therefore, to produce 18 moles of ammonia, we will need:

#?molN_2=18cancel(molNH_3)xx(1molN_2)/(2cancel(molNH_3))=9molN_2#

May 1, 2016

Compare the mole ratio between #NH_3# and #N_2#.
The mole ratio between #NH_3 : N_2# is #2:1#.

If 2 moles of #NH_3# gives you 1 mole of #N_2#,

then 18 moles of #NH_3# should give you:
[18 moles#-:# 2 #xx# 1] = 9 moles of #N_2#

Therefore 9 moles of nitrogen are required to produce 18 moles of ammonia.