Question

Ammonia reacts with O₂ to form either NO(g) or NO₂(g) according to these unbalanced equations: NH₃(g) + O₂(g) → NO(g) + H₂O(g) NH₃(g) + O₂(g) → NO₂(g) + H₂O(g) In a certain experiment 2.00 moles of NH₃(g) and 10.00 moles of O₂(g) are contained in a closed flask. After the reaction is complete, 6.75 moles of O₂(g) remains. Calculate the number of moles of NO(g) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

Answer 1

By subtracting the remaining moles of oxygen from the initial amount and applying the stoichiometry of the balanced equation, we calculate that 2.60 moles of NO(g) are produced.

Step-by-step explanation:

To calculate the number of moles of NO(g) produced when ammonia reacts with oxygen, we need to look at the reaction stoichiometry. The balanced chemical equation for the production of NO from NH3 and O2 is:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Since 2.00 moles of NH3 and 10.00 moles of O2 are initially present, and only 6.75 moles of O2 remain after the reaction, this means that 10.00 - 6.75 = 3.25 moles of O2 have reacted. From the equation, we can see that 5 moles of O2 produce 4 moles of NO, so the ratio of O2 to NO is 5:4. Using this ratio, we can calculate the moles of NO produced:

Moles of NO = (3.25 moles O2 × (4 moles NO / 5 moles O2)) = 2.60 moles of NO