N 2 ( g ) + 3 H 2 ( g ) ⟶ 2 N H 3 ( g ) In a certain reaction, 5.0 moles of hydrogen is reacted with excess nitrogen. To determine how many moles of ammonia are produced, what conversion factor should be used?

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Answer 1 · 2018-04-25T09:14:19

Here #H_2# will act as the Limiting reagent because the #N_2# is take in excess. See below :-

#N_2 + 3H_2 ⟶ 2NH_3#

On dividing by #3# , this equation can be rewritten as :-

#1/3N_2 + H_2 ⟶ 2/3NH_3#

This equation explains that for #1# mole of #H_2# the number of moles of #NH_3# formed will be #2/3#. (when #N_2# is in excess)

So for #5# moles of #H_2# the number of moles of #NH_3# formed #=2/3 xx 5 =10/3#moles

Thus the number of moles of Ammonia formed #=10/3#