Final answer:
To find out how many grams of oxygen gas are required to react with 3.01 × 10²¹ atoms of magnesium, you first convert the atoms to moles, then use the stoichiometry of the magnesium and oxygen reaction to calculate the moles of oxygen needed, and finally, multiply it by the molar mass of oxygen to get the mass in grams, which is 0.0800 g of O2.
Step-by-step explanation:
To determine how many grams of oxygen gas (O₂) are necessary to react completely with 3.01 × 10²¹ atoms of magnesium (Mg) to yield magnesium oxide (MgO), we look at the balanced chemical equation: 2Mg(s) + O₂(g) → 2MgO(s).
This equation tells us that two atoms of Mg react with one molecule of O₂ to produce two units of MgO.
First, we need to convert the number of magnesium atoms to moles using Avogadro's number (6.022 × 10²³ atoms/mol):
Number of moles of Mg = 3.01 × 10²¹ atoms ÷ (6.022 × 10²³ atoms/mol) = 5.00 × 10⁻³ mol Mg
According to the stoichiometry of the reaction, 2 moles of Mg react with 1 mole of O₂.
So for 5.00 × 10⁻³ moles of Mg, we'll need half of that or 2.50 × 10⁻³ moles of O₂.
To find out the weight of this amount of O₂, we multiply the moles by the molar mass of O₂ (32.00 g/mol):
Mass of O₂ required = 2.50 × 10⁻³ moles × 32.00 g/mol
= 0.0800 g of O₂
Therefore, 0.0800 g of oxygen gas is necessary to react with 3.01 × 10²¹ atoms of magnesium to yield magnesium oxide.