Final answer:
To produce 3.1 g of O₂, approximately 8.9 grams of NO₂ are required based on the stoichiometry of the balanced chemical reaction.
Step-by-step explanation:
The question asks how many grams of NO₂ are required to produce 3.1 g of O₂ according to the reaction 2NO₂ → 2NO + O₂. This involves using stoichiometry to calculate the molar mass of O₂ and NO₂ and then using the balanced chemical equation to find the mole ratio between O₂ and NO₂. To solve this, first, calculate the moles of O₂ produced using the molar mass of O₂, which is approximately 32 g/mol. The moles of O₂ will be 3.1 g divided by 32 g/mol, giving us approximately 0.097 moles. According to the balanced chemical equation, the mole ratio of NO₂ to O₂ is 2:1, so you'll need two moles of NO₂ for every mole of O₂ produced. Therefore, to produce the 0.097 moles of O₂, we need 0.194 moles of NO₂. Finally, using the molar mass of NO₂, which is approximately 46 g/mol, multiplying 0.194 moles by 46 g/mol will give us the number of grams of NO₂ required. This calculation results in approximately 8.9 grams of NO₂.