Question

30 POINTS! ----- How many moles of oxygen gas are needed to completely react with 145 grams of aluminum? Report your answer with 3 significant figures. ____ mole O2.

Answer 1

First a balanced reaction equation must be established:
$4Al _{(s)} + 3 O_{2} _{(g)}$     →    $2 Al_{2} O_{3}$

Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
                                                                                    =  4.83 mols

Now the mole ratio of Al : O₂ based on the equation is  4 : 3  
                                                                [4Al  + 3 O₂ → 2 Al₂O₃]

∴ if moles of Al = 4.83 moles
  then moles of O₂ = (4.83 mol ÷ 4) × 3
                              =  3.63 mol   (to  2 sig. fig.) 

Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.