Question

How many moles of O₂ would be required to generate 13.0 mol of NO₂ in the reaction below assuming the reaction has only 75.4% yield?

2 NO (g) + O₂ (g) → 2 NO₂ (g)

Answer 1

To generate 13.0 mol of NO₂ in a reaction with a 75.4% yield, approximately 17.24 mol of O₂ would be required.

Step-by-step explanation:

The reaction provided is: 2NO(g) + O₂(g) → 2NO₂(g)

Based on the balanced equation, the molar ratio between NO₂ and O₂ is 1:1. Therefore, if we want to generate 13.0 mol of NO₂, we will need 13.0 mol of O₂.

However, the question states that the reaction has a 75.4% yield. The yield is the amount of product actually obtained compared to the theoretical amount.

Therefore, we need to adjust for the yield by dividing the required moles of O₂ by the yield:

13.0 mol of O₂ ÷ 0.754 = 17.24 mol of O₂