Final answer:
The theoretical yield of NO when 50 grams of NH3 react with 110 grams of O2 is 88.13 grams, calculated by first determining the limiting reagent, which is NH3, and then using stoichiometry to find the amount of NO produced.
Step-by-step explanation:
Theoretical Yield of Nitric Oxide
To determine the theoretical yield of NO from the given reaction of NH3 with O2, we have to first write down the balanced chemical equation:
4NH3(g) + 5O2(g) → 4NO(g) + 6H₂O(l)
Next, we calculate the moles of NH3 and O2 provided:
NH3: 50g (Molar mass of NH3 = 17.03 g/mol), so moles of NH3 = 50g / 17.03 g/mol = 2.937 moles
O2: 110g (Molar mass of O2 = 32.00 g/mol), so moles of O2 = 110g / 32.00 g/mol = 3.4375 moles
Using the stoichiometry of the reaction (4 moles of NH3 react with 5 moles of O2 to produce 4 moles of NO), we can determine the limiting reagent and then the theoretical yield. Here, NH3 is the limiting reagent as it will be completely consumed first.
The stoichiometry suggests that 4 moles of NH3 will produce 4 moles of NO. Therefore, our 2.937 moles of NH3 will also produce 2.937 moles of NO. The molar mass of NO is 30.01 g/mol, leading to a theoretical yield of:
2.937 moles of NO x 30.01 g/mol = 88.13 grams of NO
This is the theoretical yield of NO that can be obtained from the given amounts of NH3 and O2.