When 8.00x10^22 molecules of ammonia react with 7.00x10^22 molecules of oxygen according to the chemical equation shown below, how many grams of nitrogen gas are produced?
4nh3(g)+3o2(> 2n2(g)+6h2o(g)

NH₃:

N = 8*10²²
NA = 6.02*10²³

n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol

O₂:

N=7*10²²
NA = 6.02*10²³

n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol

4NH₃                   +                        3O₂                      ⇒          2N₂ + 6H₂O
4mol                     :                        3mol                   :             2mol
0.133mol             :                        0.116mol           :             0,0665mol
limiting reactant

N₂:

n = 0.0665mol
M = 28g/mol

m = n*M = 0.0665mol*28g/mol = 1,862g

The answer is b metal denting