Final answer:
The stoichiometric calculation suggests 53.54 g of NO would be produced, but the reaction is not favorable, as indicated by a very small equilibrium constant (Keq = 1.0 × 10^-25), meaning actual NO production will be much lower than predicted by stoichiometry.
Step-by-step explanation:
To calculate how many grams of NO can be produced when 25.0 g of nitrogen reacts, we use stoichiometry based on the balanced chemical equation N₂(g) + O₂(g) → 2NO(g). First, we convert the mass of nitrogen to moles using its molar mass (1 mol N₂ = 28.02 g). Then we use the molar ratio from the chemical equation to determine the moles of NO produced. Finally, we convert the moles of NO to grams using the molar mass of NO (1 mol NO = 30.01 g).
Conversion of nitrogen mass to moles: 25.0 g N₂ × (1 mol N₂ / 28.02 g N₂) = 0.892 moles of N₂
Using the molar ratio, we can deduce 0.892 moles of N₂ will produce 0.892 x 2 = 1.784 moles of NO.
Conversion of NO moles to grams: 1.784 moles NO × (30.01 g NO / 1 mol NO) = 53.54 g NO
However, the reaction is not favorable under standard conditions as indicated by the extremely small equilibrium constant, Keq = 1.0 × 10^-25. This means that under normal conditions, very little NO will actually be produced when N₂ and O₂ react, regardless of stoichiometric predictions. Therefore, while the stoichiometric answer would be 53.54 g NO, the actual amount produced would be significantly less, and none of the given options (A, B, C, D) accurately depict the expected yield due to the reaction's high unfavorability.