Question

URGENT!! This is timed, PLEASE HELP!

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures, as described by the following balanced equation:
2 NH3(g) + 3 CuO(s) → 1N2(g) + 3 Cu(s) + 3 H2O(g)
How many grams of N2 are formed when 120.51 g of NH3 are reacted with excess CuO?
(Please explain using steps and show the whole process. Make sure the answer is in sig figs)

Answer 1

99.24 gm of nitrogen .

Step-by-step explanation:

molecular weight of ammonia = 17 , molecular weight of nitrogen = 28.

2 NH₃(g) + 3 CuO(s) → 1N₂(g) + 3 Cu(s) + 3 H₂O(g)

2 x 17 gm 28 gm

( 34 gm )

34 gm of ammonia forms 28 gms of nitrogen

1 gm of ammonia forms 28 / 34 gms of nitrogen

120.51 gn of ammonia forms 28 x 120.51 / 34 gms of nitrogen

28 x 120.51 / 34 gms

= 99.24 gms of nitrogen will be formed .