In the equation #N_2(g) + 3H_2(g) -> 2NH_3(g)#, 8.5 grams of hydrogen and 15.8 grams of nitrogen react to form ammonia. How many grams of ammonia are produced?

1 Answer
Dec 31, 2015

Answer:

Explanation:

Moles of #N_2# reactant #=# #(15.8*g)/(28.02*g*mol^-1)# #=# #0.564# #mol#.

Moles of #H_2# reactant #=# #(8.5*g)/(2.02*g*mol^-1)# #=# #4.21# #mol#.

Dihydrogen is the reagent in excess (clearly!). Given that dinitrogen is the limiting reagent, at most I can form #1.13# mol ammonia gas (i.e. 2 equiv with respect to dinitrogen).

If I have formed #1.13# #mol# ammonia, this is #1.13# #cancel("mol")xx17.0*g*cancel("mol"^-1)# #=# #??g#

You might already know that this is probably the most important reaction on the planet. An industrialist would have an orgasm if he could achieve such yields.