How many moles of #NH_3# are produced when 18.5 grams of #N_2# gas reacts with excess #H_2# gas?

Firstly, fully balance the chemical equation.
#N_2 + 3H_2 => 2NH_3#

Use this formula #n = m / M# where:
#n# = number of moles
#m# = mass of substance
#M# = molar mass

Secondly, you need to find out how many moles of #N_2# there are.
Use that #n = m / M# formula to do that.
#n# = ?
#m# = 18.5 grams
#M# = ?

In order to calculate what is #n#, you need to find the #M# of #N_2# first.
The molar mass can be determined right away by looking straight into the periodic table.
Molar mass of #N_2# is = 14.0 g/mol #xx# 2 = 28.0 g/mol

So,
#n# = #18.5 / 28.0# = 0.6607142857 moles [Don't round it off yet or you'll get inaccurate final answer in the end!]
#m# = 18.5 grams
#M# = 28.0 g/mol

Third step, find out what the #n# is for #NH_3#. This is done using the mole ratio.
The mole ratio between #N_2#:#NH_3# is #1:2#.

This means if 1 mole of #N_2# gives you 2 moles of #NH_3#,
then 0.6607142857 moles of #N_2# will give you:
#0.6607142857 -: 1 xx 2# = 1.32 moles of #NH_3#.

Therefore, 1.32 moles of #NH_3# are produced when 18.5 grams of #N_2# gas reacts with excess #H_2# gas.