Calculate the number of moles of NO(g) formed in the reaction mixture

Question

Ammonia reacts with O₂ to form either NO(g) or NO₂(g) according to the following UNBALANCED equations: NH₃(g) + O₂(g) → NO(g) + H₂O(g) NH₃(g) + O₂(g) → NO₂(g) + H₂O(g) In one experiment 6.00 moles of NH₃(g) and 15.00 moles of O₂(g) were placed in a closed flask and allowed to react. After the reaction was complete, 5.15 moles of O₂(g) remained and all of the NH₃(g) had reacted. Calculate the number of moles of NO(g) formed in the reaction mixture

Answer 1

Let me try again, after giving this some additional thought.

(1) 4NH3 + 5O2 --> 4NO + 6H2O

(2) 4NH3 + 7O2 --> 4NO2 + 6H2O

1 mol NH3 produces 1 mol NO

1 mol NH3 produces 1 mol NO2

Let X = moles of NH3 that produces X moles of NO

Then (8-4x) = moles of NH3 used to produce NO2

Total moles O2 used = 15.00 - 5.15 = 9.85 moles O2 used

(1) 4NH3 + 5O2 --> 4NO + 6H2O

(X mol NH3) (5 moles O2 / 4 moles of NH3) = 1.25X mol O2 used in equation (1)

(2) 4NH3 + 7O2 --> 4NO2 + 6H2O

( 8 - 4X moles of NH3) (7 moles O2 / 4 moles of NH3) = (8 - 4X) (1.75 mol O2) moles O2 used in equation (2)

1.25X mol O2 + (8 - 4X) (1.75 mol O2) = 9.85 mol O2 used

1.25X + 14 - 7X = 9.85

X = 0.722 moles NH3 = 0.722 moles NO