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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.

Respuesta :

transitionstate
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

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