Question

An important reaction sequence in the industrial production of nitric acid is the following:


`N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)`

`4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(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?
A) 12.5 mol O₂
B) 20.0 mol O₂
C) 25.0 mol O₂
D) 50.0 mol O₂
E) 100. mol O₂

Answer 1

25.0 mol O₂ are required in the second reaction

Step-by-step explanation:

N₂ (g) + 3H₂ (g) → 2NH₃ (g)

4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6 H₂O(l)

Molar ratio in first reaction is 1:2

For every mol of N₂. I make 2 moles of ammonia. If I have 20 moles of N₂, i'm going to get, 40 moles of ammonia.

In the second reaction, molar ratio between products is 4:5.

If I obtained 40 moles of ammonia in first step, let's prepare the rule of three.

4 moles of ammonia react with 5 moles of O₂

40 moles of ammonia react with ( 40.5) /4 = 25moles