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The amount of Al required will be 15.77 grams

Stoichiometric problem

First, the equation of the reaction:

[tex]2Al +3 FeO -- > Al_2O_3 + 3Fe[/tex]

The mole ratio is 2:3.

Mole of 63.0 g of FeO = 63/71.84 = 0.8769 moles

Equivalent moles of Al = 0.8769 x 2/3 = 0.5846 moles

Mass of 0.5846 moles Al = 0.5846 x 26.98 = 15.77 grams

More on stoichiometric problems can be found here: https://brainly.com/question/14465605

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CintiaSalazar

Taking into account the reaction stoichiometry, 13.28 grams of Al is required to react with 63 grams of FeO.

Reaction stoichiometry

In first place, the balanced reaction is:

2 Al + 3 FeO → 3 Fe + 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: 2 moles
  • FeO: 3 moles
  • Fe: 3 moles
  • Al₂O₃ : 1 mole

The molar mass of the compounds is:

  • Al: 27 g/mole
  • FeO: 71.85 g/mole
  • Fe: 55.85 g/mole
  • Al₂O₃ : 102 g/mole

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

  • Al: 2 moles ×27 g/mole= 54 grams
  • FeO: 3 moles ×71.85 g/mole= 215.55 grams
  • Fe: 3 moles ×55.85 g/mole= 167.55 grams
  • Al₂O₃ : 1 mole ×102 g/mole= 102 grams

Mass of Al required

The following rule of three can be applied: If by reaction stoichiometry 215.55 grams of FeO react with 54 grams of Al, 53 grams of FeO react with how much mass of Al?

[tex]mass of Al= \frac{53 grams of FeOx54 grams of Al}{215.55 grams of FeO}[/tex]

mass of Al= 13.28 grams

Finally, 13.28 grams of Al is required to react with 63 grams of FeO.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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