How many liters of oxygen at 0 ∘c and 760 mmhg (stp) are required to completely react with 4.2 g of aluminum?

Reaction between aluminium and oxygen can be represented as:

4Al(s) + 3O₂(g) >2Al₂O₃(s)

Number of moles of aluminium = Mass of aluminium/ Molar mass of aluminium

=4.2g /26.981 g / mol

=0.155 mol

The equation shows that these react with 0.155 * 3/4 moles of O2 = 0.11 moles of O2.

1 mole of O2 at STP occupies 22.4 liter

volume of O2 = 0.11 moles ×22.4 liters / mol

= 2.61 liter of O2.

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