Grams of #N_2# reacts with 84.29 grams of #H_2# in the chemical equation: #N_2 + H_2-> NH_3# How do you balance the chemical equation and determine the limiting reactant? How much #NH_3# is produced?

1 Answer
Jan 6, 2016

Answer:

The stoichiometric equation is:

#1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)#

Explanation:

Three equiv of hydrogen react with 1 equiv of nitrogen to give one equiv ammonia. Of course, I could double this reaction to remove the 1/2 coefficients, but the stoichiometry will remain the same.

If you have 84.29 g dihydrogen, there are #(84.29*cancelg)/(2.02*cancelg*mol^-1)# #=# #41.7# #mol# #H_2#. Stoichiometry requires 27.8 mol dinitrogen (#=# #?? g#). I think you have omitted some of the starting conditions in the problem