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The chemical equation below shows the formation of aluminum oxide (Al2O3) from aluminum (Al) and oxygen (O2). 4Al 3O2 Right arrow. 2Al2O3 The molar mass of O2 is 32. 0 g/mol. What mass, in grams, of O2 must react to form 3. 80 mol of Al2O3? 60. 8 81. 1 122 182.

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CintiaSalazar

Taking into account the stoichiometry of the reaction, 182.4 grams of O₂ must react to form 3.80 mol of Al₂O₃.

The balanced reaction is:

4 Al + 3 O₂ → 2 Al₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Al: 4 moles
  • O₂: 3 moles
  • Al₂O₃: 2 moles

The molar mass of each compound is:

  • Al: 27 g/mole
  • O₂: 32 g/mole
  • Al₂O₃: 102 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:  

  • Al: 4 moles× 27 g/mole= 108 grams
  • O₂: 3 moles× 32 g/mole= 96 grams
  • Al₂O₃: 2 moles× 102 g/mole= 204 grams

Then it is possible to apply the following rule of three:  If by reaction stoichiometry 2 moles of Al₂O₃ are formed from 96 grams of O₂, 3.80 moles of Al₂O₃ are formed from how much mass of O₂?

[tex]mass of O_{2}=\frac{3.80 moles of Al_{2} O_{3} x96 grams of O_{2} }{2 moles of Al_{2} O_{3} }[/tex]

mass of O₂= 182.4 grams

In summary, 182.4 grams of O₂ must react to form 3.80 mol of Al₂O₃.

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