Final answer:
To produce 85.9 moles of nitrogen monoxide, 42.95 moles of elemental oxygen gas are required, based on the stoichiometric ratio from the balanced chemical reaction.
Step-by-step explanation:
The student is asking how many moles of elemental oxygen gas are required to produce 85.9 moles of nitrogen monoxide (NO). This question pertains to stoichiometry, a concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. According to the balanced chemical equation for the formation of nitrogen monoxide:
The equation shows that 1 mole of nitrogen (N₂) reacts with 1 mole of oxygen (O₂) to produce 2 moles of NO. Therefore, to produce 85.9 moles of NO, half that amount in moles of oxygen gas is needed. This is because the molar ratio of NO to O₂ in the balanced equation is 2:1. Using simple division:
85.9 moles NO × (1 mole O₂ / 2 moles NO) = 42.95 moles O₂
Thus, 42.95 moles of elemental oxygen gas are required to produce 85.9 moles of nitrogen monoxide.