An important reaction sequence in the industrial production of nitric acid is the following: N2(g) + 3H2(g) → 2NH3(g) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) Starting from 20.0 mol of nitrogen gas in the first reaction, how many moles of oxygen gas are required in the second one?

The answer to your question is 50 moles of O₂

Explanation:

Balanced Chemical reactions

1.-                 N₂(g)  +  3H₂ (g)   ⇒   2NH₃ (g)

2.-                4NH₃ (g) + 5O₂(g)  ⇒  4NO (g)  +  6H₂O (l)

moles of N₂(g) = 20 moles

moles of O₂(g) = ?

Process

1.- Calculate the moles of NH₃

                     1 mol of N₂ 2 moles of NH₃

                   20 moles of N₂  x

                     x = (20 x 2) / 1

                     x = 40 moles of NH₃

2.- Calculate the moles of O₂

                4 moles of NH₃ 5 O₂

               40 moles of NH₃  x

                    x = (40 x 5) / 4

                    x = 200 / 4

                    x = 50 moles of O₂