Considering the reaction stoichiometry, 51.22 grams of Fe are produced from 65.9 g of FeO.
The balanced reaction is:
3 FeO + 2 Al → 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:
- FeO: 3 moles
- Al: 2 moles
- Fe: 3 moles
- Al₂O₃: 1 mole
The mass molar of the elements and compounds is:
- FeO: 71.85 g/mole
- Al: 27 g/mole
- Fe: 55.85 g/mole
- Al₂O₃: 102 g/mol
Then, by reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:
- FeO: 3 moles× 71.85 g/mole= 215.55 grams
- Al: 2 moles× 27 g/mole= 54 grams
- Fe: 3× 55.85 g/mole= 167.55 grams
- Al₂O₃: 1× 102 g/mol= 102 grams
Then you can apply the following rule of three: if by stoichiometry 215.55 grams of FeO produce 167.55 grams of Fe, 65.9 grams of FeO produces how much mass of Fe?
[tex]mass of Fe=\frac{65.9 grams of FeOx167.55 grams of Fe}{215.55 grams of FeO}[/tex]
mass of Fe= 51.22 grams
Finally, 51.22 grams of Fe are produced from 65.9 g of FeO.
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