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Aluminum oxidizes according to the following equation: 4 Al + 3 O2 -> 2 AL2 O3....

Aluminum oxidizes according to the following equation: 4 Al + 3 O2 -> 2 AL2 O3.

If 0.048 mol of Al is placed into a container with 0.030 mol O2, what are the limiting and excess reactants? Based on the previous question, how many moles of excess reactant remain?
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Answer #1

Given that 4 A1+302 2 Al2 O3 mole of Al=0.048 mol mole of O2 = 0.030 mol . For limiting Reagent Al 0.048 = 0.012 02 + 0.030 =Please give better rating

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