Final answer:
The mass of O₂ needed to react with 8.10 g of ethanol is approximately 16.9 g, based on stoichiometric calculations which is not among the given answer choices.
Step-by-step explanation:
To determine how many grams of O₂ are needed to react with 8.10 g of ethanol, C₂H₆O, we must first use the balanced chemical equation: C₂H₆O(aq) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). Next, we find the molar mass of ethanol (46.07 g/mol) and oxygen (32.00 g/mol). Then, we use stoichiometry to calculate the moles of ethanol, which is 8.10 g / 46.07 g/mol ≈ 0.176 moles. The chemical equation shows that 1 mole of ethanol reacts with 3 moles of O₂, so for 0.176 moles of ethanol, we need 0.176 moles × 3 ≈ 0.528 moles of O₂. Finally, we find the mass of this quantity of oxygen: 0.528 moles × 32.00 g/mol ≈ 16.9 g which is not listed among the answer choices, indicating a possible error in the options given.