How many grams of Al are needed to react with 63.0 g of FeO
2 answers:
Taking into account the reaction stoichiometry, 13.28 grams of Al is required to react with 63 grams of FeO.
<h3>Reaction stoichiometry</h3>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
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?
<u><em>mass of Al= 13.28 grams</em></u>
Finally, 13.28 grams of Al is required to react with 63 grams of FeO.
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The amount of Al required will be 15.77 grams
<h3>Stoichiometric problem</h3>First, the equation of the reaction:
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: brainly.com/question/14465605
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