Calculate the volume of oxygen required to burn 3 g of aluminum

The reaction of aluminum with oxygen proceeds according to the following chemical formula:

4Al + 3O2 = 2Al2O3;

That is, for 4 moles of aluminum, 3 moles of oxygen are needed.

The molar mass of aluminum is 27 grams / mol

The number of moles of a substance in 3 grams of aluminum is 3/27 = 0.11111 mol;

The required amount of oxygen will be 0.1111 * 3/4 = 0.0833 mol;

Under normal conditions 1 mole of ideal gas takes up a volume of 22.4 liters.

The amount of oxygen equal to 0.0833 mol occupies a volume under normal conditions equal to 0.0833 x 22.4 = 1.8659 liters.

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