aluminum metal combines with 22.40 liters of oxyen gas to form solid aluminum oxide

There must be a question here but I don't see it.

To determine the mass of aluminum oxide formed, you need to know the balanced chemical equation for the reaction between aluminum and oxygen. The balanced equation would be:

4 Al + 3 O2 -> 2 Al2O3

From the equation, you can see that 4 moles of aluminum react with 3 moles of oxygen to form 2 moles of aluminum oxide.

First, determine the number of moles of oxygen gas provided. To do this, you need to know the volume of the gas at STP (Standard Temperature and Pressure) conditions since you are given the volume of oxygen gas in liters. At STP, 1 mole of any gas occupies 22.4 liters.

Given that you have 22.40 liters of oxygen gas, you can calculate the number of moles using the following calculation:

moles of oxygen gas = volume of gas / molar volume at STP
= 22.40 L / 22.4 L/mol
= 1 mole

Now that you have determined that you have 1 mole of oxygen gas, you can use the stoichiometry of the equation to find the number of moles of aluminum that would react. According to the equation, 3 moles of oxygen gas are required to react with 4 moles of aluminum. So, the number of moles of aluminum would be:

moles of aluminum = (moles of oxygen) x (moles of Al / moles of O2)
= 1 mole x (4 moles Al / 3 moles O2)
= 1.33 moles Al (rounded to two decimal places)

The molar mass of aluminum (Al) is 26.98 g/mol and the molar mass of aluminum oxide (Al2O3) is 101.96 g/mol. Using these molar masses, you can determine the mass of aluminum oxide formed:

mass of Al2O3 = (moles of Al2O3) x (molar mass of Al2O3)
= (2 moles) x (101.96 g/mol)
= 203.92 g

Thus, the mass of solid aluminum oxide formed is 203.92 grams.