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.