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.