When 8.00 * 1022 molecules of ammonia react with 7.00 * 1022 molecules of oxygen according to the chemical equation shown below, how many grams of nitrogen gas are produced?

4 NH3 (g) + 3 O2 (g) → 2 N2 (g) + 6 H2O (g)

According to the equation: 4 moles of NH3 requires 3 moles of Oxygen

1 mole = 6.02 x 10^23

X moles = 8 x 10^22

X=0.13 moles of NH3

and for Oxygen

X = (7x10^22) / (6.02x10^23)

X=0.116 of Oxygen therefore Ammonia is the limiting factor.

4 moles NH3: 2 moles N2

0.13 moles: X

X=0.065 moles of Nitrogen gas will be formed.