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)

99.24 gm of nitrogen .

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 .

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.