Question

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

Answer 1

 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!