Aluminum reacts with oxygen to produce aluminum oxide which can be used as an adsorbent, desiccant or catalyst for organic reactions. a mixture of 82.49 g of aluminum and 117.65 g of oxygen is allowed to react. identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

The balance chemical equation is as follow,

                                        2 Al  +  3 O₂   →   Al₂O₃

Aluminium is the Limiting Reagent,
As,
                                107.92 g Al required  =  96 g of O₂
Then,
                          82.49 g of Al will require  =  X g of O₂
Solving for X,
                                    X  =  (82.49 g × 96 g) ÷ 107.92 g

                                    X  =  73.37 g of O₂
But,
     We are provided with 117.65 g of O₂, So, it is provided in excess and 44.28 g of it will remain unreacted.

Solving for Amount of Al₂O₃ formed,
As,
                        107.92 g of Al produced  =  203.92 g of Al₂O₃
Then,
                      82.49 g of Al will produce  =  X g of Al₂O₃
SOlving for X,
                                          X  =  (82.49 g ×  203.92 g) ÷ 107.92

                                          X  =  155.86 g of Al₂O₃

Do u have the n? or maybe the m?