Final answer:
To find the grams of oxygen gas needed to react with 3.01 × 10²¹ atoms of magnesium, we use stoichiometry. However, the calculation gives 0.0800 g of oxygen gas, which does not match the provided options.
Step-by-step explanation:
To determine how many grams of oxygen gas are necessary to react completely with 3.01 × 10²¹ atoms of magnesium to yield magnesium oxide, we need to use stoichiometry based on the balanced chemical equation for the reaction of magnesium with oxygen:
2Mg + O₂ → 2MgO
First, we calculate the number of moles of magnesium atoms using Avogadro's number:
3.01 × 10²¹ atoms Mg × (1 mole Mg / 6.022 × 10²³ atoms) = 5.00 × 10⁻³ moles Mg
According to the balanced equation, 1 mole of O₂ reacts with 2 moles of Mg. Therefore, the moles of O₂ needed for the reaction are half the moles of Mg:
5.00 × 10⁻³ moles Mg × (1 mole O₂ / 2 moles Mg) = 2.50 × 10⁻³ moles O₂
Now, we convert moles of oxygen gas to grams using the molar mass of O₂ (32.00 g/mol):
2.50 × 10⁻³ moles O₂ × (32.00 g O₂ / 1 mole O₂) = 0.0800 g O₂
Since 0.0800 g of O₂ is needed and this is not an option in the multiple-choice answers, there seems to be a discrepancy. Please double-check the provided answer choices and the details of the question to ensure there is no mistake.