Answer: 33.0 grams
Step-by-step explanation:
To solve this problem, we need to use stoichiometry to find the amount of NO that will be produced from the given amounts of NH₃ and O₂, and then convert that amount to mass.
First, we need to determine which reactant is limiting, i.e. which one will be completely used up in the reaction. To do this, we can compare the number of moles of NH₃ and O₂ that are present with the stoichiometric ratio of NH₃ to O₂ in the reaction:
4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)
From the balanced equation, we can see that the ratio of NH₃ to O₂ required for complete reaction is 4:5. Therefore, we need to convert the given masses of NH₃ and O₂ to moles, and compare them using this ratio:
Moles of NH₃ = 25.0 g / 17.03 g/mol = 1.47 mol
Moles of O₂ = 43.9 g / 32.00 g/mol = 1.37 mol
NH₃:O₂ = 1.47 mol : 1.37 mol = 1.07 : 1
Since the ratio of NH₃ to O₂ is less than the stoichiometric ratio of 4:5, we can conclude that O₂ is the limiting reactant.
Now we can use the amount of O₂ to calculate the amount of NO that will be produced:
Moles of O₂ used = 1.37 mol
Moles of NO produced = (4 mol NO / 5 mol O₂) x 1.37 mol O₂ = 1.10 mol NO
Finally, we can convert the amount of NO to mass using its molar mass:
Mass of NO = 1.10 mol x 30.01 g/mol = 33.0 g
Therefore, 33.0 grams of NO will be formed when 25.0 g of NH₃ and 43.9 g of O₂ react in the given reaction.