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

The two reactions are competing for the O2 and so some NO will form and some NO2 will form. Our job is to figure out how much NO is formed.

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 (2 - X) = moles of NH3 that produced the (2 - X) moles of NO2

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

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

1.25X mol O2 + (2 - X) (1.75 mol O2) = 9.85 mol O2 used

1.25X + 3.5 - 1.75X = 9.85 mol O2

1.25X - 1.75X = 9.85 - 3.5

Solving for x results in a negative number. Not sure where I went wrong here, but I think this is the correct approach to answer this question. You should definitely check my math and my reasoning. X should be positive and should be the moles NH3 used to produce NO in eq (1) and that would also be the moles of NO.