The common Haber process equation...
#N_2(g)+3H_2(g)stackrel(450^@C)stackrel(Fe \ "catalyst")stackrel(200 \ "atm")rightleftharpoons2NH_3(g)#
We see that the mole ratio between #N_2# and #NH_3# is #1:2#. So, for every mole of nitrogen gas used, two moles of ammonia gas are made.
We have #50 \ "g"# of nitrogen gas here, so we need to find the number of moles of nitrogen.
#N_2# has a molar mass of #28 \ "g/mol"#. So here, there exist
#(50color(red)cancelcolor(black)"g")/(28color(red)cancelcolor(black)"g""/mol")~~1.79 \ "mol"#
From here, we would produce #1.79*2=3.58# moles of ammonia.
Ammmonia #(NH_3)# has a molar mass of #17.031 \ "g/mol"#. So here, we would produce
#3.58color(red)cancelcolor(black)"mol"*(17.031 \ "g")/(color(red)cancelcolor(black)"mol")~~61 \ "g"# of #NH_3#.
