What is the percent yield for the reaction between 15.8 g of #NH_3# and excess oxygen to produce 21.8 g of #NO# gas and water?

1 Answer
Apr 23, 2016

Answer:

#2NH_3(g) + 5/2O_2(g) rarr 2NO(g) + 3H_2O(l)#. The yield is approx. #70%#.

Explanation:

I assume that #15.8*g# of ammonia reacts with #O_2# to give #21.8*g# #NO#, and water. Ammonia is the limiting reagent.

#"Moles of ammonia"# #=# #(15.8*g)/(17.03*g*mol^-1)# #=# #0.928*mol#

#"Moles of "NO# #=# #(21.8*g)/(30.01*g*mol^-1)# #=# #0.726*mol#

Since the ratio between #NH_3# and #NO# is #1:1#, the yield is #(0.726*mol)/(0.928*mol)# #xx100%# #=# #??%#