N2 Â Â +Â Â 3 H2 -------------------> 2
NH3
1mol     3 mol
         Â
          Â
2 mol
5 mol    11
mol                           Â
?
here limiting reagent is H2. so NH3 formed based on that.
3 mol H2 -------------- 2 mol NH3
11 mol H2 -------------- ?? NH3
moles of NH3 produced = 11 x 2 / 3
                                        Â
= 7.33 mol NH3
we need only 3.67 mol of N2. here we have 5 mol of N2 . so N2 is
left over
moles of N2 remain = 5 - 3.67 = 1.33 mol N2
moles of H2 = 0 . it all consumed
