Final answer:
To completely react with 9.30 moles of aluminum, 223.2 grams of oxygen gas (O₂) are needed.
Step-by-step explanation:
To determine how many grams of oxygen gas (O₂) are needed to completely react with 9.30 moles of aluminum, we can use the balanced chemical equation for the reaction between aluminum and oxygen:
4 Al + 3 O₂ → 2 Al₂O₃
The stoichiometric ratio tells us that 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide. Therefore, to find the number of moles of oxygen gas needed, we can set up a proportion:
4 moles Al / 3 moles O₂ = 9.30 moles Al / x moles O₂
Solving for x, we get x = (3/4) * 9.30 moles O₂ = 6.975 moles O₂
Next, we can convert the moles of oxygen gas to grams using the molar mass of O₂:
Molar mass of O₂ = 32.00 g/mol
Grams of O₂ = 6.975 moles O₂ * 32.00 g/mol = 223.2 g
Therefore, 223.2 grams of oxygen gas are needed to completely react with 9.30 moles of aluminum.