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

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

what are the two models

explanation: