Final answer:
To completely react with 9.30 moles of aluminum, you would need 223.2 grams of oxygen gas (O2).
Step-by-step explanation:
To calculate the grams of oxygen gas necessary to react completely with a given amount of aluminum, you need to use the stoichiometry of the reaction. The balanced equation for the reaction between aluminum and oxygen gas (O2) to form aluminum oxide (Al2O3) is:
4Al + 3O2 → 2Al2O3
From the balanced equation, we can see that 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide. Therefore, the molar ratio between aluminum and oxygen gas is 3:4.
Given that you have 9.30 moles of aluminum, you can use the molar ratio to calculate the moles of oxygen gas required:
9.30 moles Al * (3 moles O2 / 4 moles Al) = 6.975 moles O2
To convert moles of oxygen gas to grams, you can use the molar mass of oxygen (32.00 g/mol):
6.975 moles O2 * 32.00 g/mol = 223.2 grams of oxygen gas
Therefore, you would need 223.2 grams of oxygen gas (O2) to completely react with 9.30 moles of aluminum.