WHAT IS THE NUMBER OF MOLES OF ALUMINIUM OXIDE FORMED WHEN 6.75g OF ALUMINIUMIS MADE TO REACT WITH 8g OF OXYGEN.IS I/8MOLE THE RIGHT ANSWER?
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yes the answer is 1/8
4 Al + 3O2 ---> 2Al2O3
so limiting reagent is Al so Al2O3 is half the moles of Al so ans is 1/8
4 Al + 3O2 ---> 2Al2O3
so limiting reagent is Al so Al2O3 is half the moles of Al so ans is 1/8
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