In the reaction of aluminum metal and oxygen to make aluminum oxide, how many grams of oxygen gas will react with 2.2 moles aluminum metal?

52.8 g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4Al + 3O2 —> 2Al2O3

From the balanced equation above,

4 moles of Al reacted with 3 moles of O2 to produce 2 moles of Al2O3

Next, we shall determine the number of mole of O2 needed to react with 2.2 moles of Al. This can be obtained as follow:

From the balanced equation above,

4 moles of Al reacted with 3 moles of O2.

Therefore, 2.2 moles of Al will react with = (2.2 × 3)/4 = 1.65 moles of O2.

Thus, 1.65 moles of O2 is needed for the reaction.

Finally, we shall determine the mass of O2 needed as shown below:

Mole of O2 = 1.65 moles

Molar mass of O2 = 2 × 16= 32 g/mol

Mass of O2 =?

Mole = mass/Molar mass

1.65 = mass of O2 /32

Cross multiply

Mass of O2 = 1.65 × 32

Mass of O2 = 52.8 g

Therefore, 52.8 g of O2 is needed for the reaction.