Question

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?

Answer 1

Answer:

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