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48g of 02 produce how many grams of Al2O3

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CintiaSalazar

Taking into account the reaction stoichiometry,  102 grams of Al₂O₃ are formed when 48 grams of O₂ react.

Reaction stoichiometry

In first place, 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 the compounds 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

Mass of Al₂O₃ formed

The following rule of three can be applied: if by reaction stoichiometry 96 grams of O₂ form 204 grams of Al₂O₃, 48 grams of O₂ form how much mass of Al₂O₃?

[tex]mass of Al_{2} O_{3} =\frac{48 grams of O_{2} x204 grams of Al_{2} O_{3}}{96 grams of O_{2}}[/tex]

mass of Al₂O₃= 102 grams

Finally, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.

Learn more about the reaction stoichiometry:

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