Final Answer:
a. 33.2 g NO produced when 25.0 g nitrogen reacts.
b. 100.0 g NO produced when 25.0 g oxygen reacts.
c. Predicted NO amount is 33.2 g when 25.0 g nitrogen reacts with 25.0 g oxygen; based on limiting reactant (nitrogen) stoichiometry.
Step-by-step explanation:
a. To determine the amount of NO produced when 25.0 g of nitrogen reacts, we use the molar ratio between nitrogen and NO in the balanced equation. The molar mass of nitrogen is 28.02 g/mol, and the molar mass of NO is 30.01 g/mol. Using these values, we find that 25.0 g of nitrogen produces 33.2 g of NO.
b. Similarly, when 25.0 g of oxygen reacts, we use the molar ratio between oxygen and NO. The molar mass of oxygen is 32.00 g/mol. Based on the balanced equation, 25.0 g of oxygen produces 100.0 g of NO.
c. Since nitrogen is the limiting reactant, the amount of NO produced is limited by the initial amount of nitrogen. Therefore, the predicted amount of NO is 33.2 g.
d. The reasoning behind this prediction lies in the stoichiometry of the reaction. The balanced equation indicates that 1 mole of nitrogen produces 2 moles of NO. By converting the moles of nitrogen to moles of NO using the molar ratio, and then to grams, we obtain the predicted amount of NO, which is 33.2 g. This calculation assumes complete reaction and adherence to the stoichiometric proportions outlined in the balanced chemical equation.