For the reaction of ammonia with molecular oxygen, forming nitrogen monoxide and water. How do you calculate the theoretical yield of #NO# when 17.0g #NH_3# reacts with 16.0g #O_2#?

1 Answer
Oct 5, 2016

Answer:

Well first you write a stoichiometrically balanced equation:

#2NH_3(aq) + 5/2O_2(g) rarr 2NO(g) +3H_2O(l)#

Explanation:

Given stoichiometric oxygen, ammonia gives an equimolar quantity of #NO#.

#"Moles of ammonia"# #=# #(17.0*g)/(17.03*g)# #~=# #1*mol#

#"Moles of dioxygen"# #=# #(16.0*g)/(32.03*g)# #~=# #0.50*mol#

Clearly, dioxygen is in deficiency, and only #4/5xx0.5*mol=0.40*mol# #NH_3# will react.

And thus only #0.40*mol# #NO# will be produced, i.e. #0.40*molxx30.01*g*mol^-1# #=# #12.0*g# #NO#.