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
anor277
Dec 31, 2015

Find the limiting reagent.

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.

Related questions
See all questions in Equation Stoichiometry
Impact of this question
2204 views around the world
You can reuse this answer
Creative Commons License