Respuesta :

CintiaSalazar

Answer:

0.472 grams of aluminum oxide would be produced from the reaction of 0.25 grams of aluminum with excess oxygen gas.

Explanation:

First of all, you should know that the balanced reaction that occurs:

4 Al + 3 O₂ → 2 Al₂O₃

Then, by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts are produced in moles:

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

Being:

  • Al: 27 g/mole
  • O: 16 g/mole

The molar mass of the compounds participating in the reaction is:

  • Al: 27 g/mole
  • O₂: 2*16 g/mole= 32 g/mole
  • Al₂O₃: 2*27 g/mole + 3*16 g/mole= 102 g/mole

Then, by stoichiometry of the reaction, the following amounts of mass react and are produced:

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

 

Now it is possible to apply the following rule of three: if 108 grams of Al produce 204 grams of Al₂O₃, 0.25 grams of aluminum how much mass of Al₂O₃ will it produce?

[tex]mass of Al_{2} O_{3} =\frac{0.25 grams of Al*204 grams of Al_{2} O_{3} }{108 grams of Al}[/tex]

mass of Al₂O₃=0.472 grams

0.472 grams of aluminum oxide would be produced from the reaction of 0.25 grams of aluminum with excess oxygen gas.

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