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.
