How many grams of ammonia, #NH_3#, would be formed from the complete reaction of 4.50 moles hydrogen, #H_2# in the reaction #N_2 + 3H_2 -> 2NH_3#?

1 Answer
Aug 1, 2016

Answer:

Approx. #54*g" ammonia"#

Explanation:

You have almost solved this problem yourself, because you have written a stoichiometrically balanced equation, which establishes equivalent mass:

#N_2(g) + 3H_2(g)rightleftharpoons 2NH_3(g)#

This tells us that #28*g# of dinitrogen combines with #6*g# dihydrogen to give #34*g# of ammonia. I take it that you know how I got these equivalent masses (if no, I am happy to explain).

You have #4.5*mol# dihydrogen, and thus, at most I can form #4.5*mol*H_2xx(2*mol" ammonia")/(3*mol" dihydrogen")# #=# #3*mol*" ammonia"#.

#"Mass of ammonia"# #=# #"moles "xx" molar mass"# #=# #3*molxx17.0*g*mol^-1# #=# #??g#