Ammonia reacts with oxygen to form nitric oxide and water vapor:
4NH₃ + 5O₂ $\rarr$ 4NO + 6H₂O
When 40.0 g NH₃ and 50.0 g O₂ are allowed to react, which is the limiting reagent

Question

Ammonia reacts with oxygen to form nitric oxide and water vapor:
4NH₃ + 5O₂

\rarr

4NO + 6H₂O
When 40.0 g NH₃ and 50.0 g O₂ are allowed to react, which is the limiting reagent

Quizplus · Accepted Answer

To determine the limiting reagent, we need to calculate the amount of product that can be produced from each reactant. 1. From the given equation, we see that the stoichiometric ratio of NH₃ to NO is 4:4 or 1:1. Therefore, 40.0 g NH₃ can produce (4/1) x (40.0 g) = 160 g of NO. 2. Similarly, the stoichiometric ratio of O₂ to NO is 5:4. Therefore, 50.0 g O₂ can produce (4/5) x (50.0 g) = 40.0 g of NO. Since O₂ can produce less NO than NH₃, it is the limiting reagent. Therefore, the correct answer is B.

Ammonia Reacts with Oxygen to Form Nitric Oxide and Water →\rarr→

Ammonia Reacts with Oxygen to Form Nitric Oxide and Water $\rarr$