Ammonia reacts with oxygen to form nitric oxide and water vapor:
4NH₃ + 5O₂ $\rarr$ 4NO + 6H₂O
When 20.0 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 20.0 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 formed from each reactant and see which one produces the least amount of product. Using the balanced chemical equation, we can see that each mole of NH₃ produces 1 mole of NO, while each mole of O₂ produces 1.25 moles of NO. From the given masses, we can calculate the moles of each reactant: moles of NH₃ = 20.0 g / 17.03 g/mol = 1.18 mol moles of O₂ = 50.0 g / 32.00 g/mol = 1.56 mol Therefore, we can see that NH₃ produces the least amount of product (1.18 moles of NO), and is therefore the limiting reagent. Answer choice A is correct.

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

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