Final answer:
The volume of oxygen gas required to react with 23.0 g of Al is 34.16 L.
Step-by-step explanation:
To find the volume of oxygen gas required in the reaction, we need to use stoichiometry and the ideal gas law equation: PV = nRT. First, let's find the number of moles of Al using its molar mass (26.98 g/mol):
Moles of Al = mass of Al / molar mass of Al = 23.0 g / 26.98 g/mol = 0.853 mol
According to the balanced equation, 4 moles of Al react with 3 moles of O2 to form 2 moles of Al2O3. Therefore:
Moles of O2 = (4/3) * moles of Al = (4/3) * 0.853 mol = 1.137 mol
Finally, we can use the ideal gas law equation to find the volume of oxygen gas:
V = (n * R * T) / P = (1.137 mol * 0.0821 L·atm/mol·K * 298 K) / 0.87 atm = 34.16 L