Given the reaction: 4al(s)+3o2(g)==2al2o3(s) what is the minimum number of grams of oxygen gas required to produce 1.00 mole of aluminum oxide

 4 Al(s)+3 O₂(g) > 2 Al₂O₃(s) 

3 moles O₂ > 2 moles Al₂O₃
? moles O₂ > 1.00 mole Al₂O3

O₂ = 1.00 * 3 / 2

O₂ = 3 / 2

= 1.5 moles of O₂

Molar mass O₂ = 32.0 g

1.5 * 32.0 = 48 g of O₂

hope this helps!

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