What is minimum volume of oxygen required to react with 42.5 g of aluminum in the synthesis of aluminum oxide at stp?

26.833 liters

Aluminum oxide has a formula of Al₂O₃, which means for every mole of aluminum used, 1.5 moles of oxygen is required (3/2 = 1.5).

Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.

Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.

Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815

Finally, you need to calculate the volume of 1.1815 moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,

1.1815 * 22.7 = 26.8 liters of oxygen gas.