The number of moles of oxygen required to completely react with 54 g of aluminum is 1.5 moles.
Number of moles of the reacting aluminum
moles = reacting mass/molar mass
moles = 54/27
moles = 2 moles
From the given reaction of oxygen and aluminum;
4Al + 30₂ → 2Al₂O₃
4 moles of Al ----------> 3 moles of oxygen
2 moles of AL --------> ? moles of oxygen
= (2 x 3)/4
= 1.5 moles
Thus, the number of moles of oxygen required to completely react with 54 g of aluminum is 1.5 moles.
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