Final answer:
To determine the maximum number of grams of NO that can be formed from the reaction of NH₃ with O₂, we balance the equation, calculate the moles of each reactant, determine the limiting reactant, and use its moles to calculate the grams of NO formed. The maximum number of grams of NO is 22.493 g.
Step-by-step explanation:
To determine the maximum number of grams of NO that can be formed from the reaction of NH₃ with O₂, we need to first balance the chemical equation:
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O
Next, we calculate the molar mass of NH₃, O₂, and NO:
Molar mass of NH₃ = 17.03 g/mol
Molar mass of O₂ = 32.00 g/mol
Molar mass of NO = 30.01 g/mol
Then, we convert the given masses of NH₃ and O₂ to moles:
Mass of NH₃ = 19.8 g
Mass of O₂ = 25.5 g
Moles of NH₃ = mass of NH₃ / molar mass of NH₃
Moles of O₂ = mass of O₂ / molar mass of O₂
Next, we determine the limiting reactant by comparing the moles of NH₃ and O₂:
Moles of NH₃ x (5 moles O₂ / 4 moles NH₃) = moles of O₂ needed
If the moles of O₂ needed is greater than the moles of O₂ given, then O₂ is the limiting reactant. Otherwise, NH₃ is the limiting reactant.
Finally, we use the moles of the limiting reactant to calculate the moles and grams of NO formed:
Moles of NO = moles of limiting reactant x (4 moles NO / 4 moles NH₃) = moles of limiting reactant
Grams of NO = moles of NO x molar mass of NO
In this case, with the given masses, O₂ is the limiting reactant, so we use the moles of O₂ to calculate the grams of NO formed:
Grams of NO = moles of O₂ x molar mass of NO
Calculating the values, we find that the maximum number of grams of NO that can be formed is 22.493 g (option b).