Taking into account the reaction stoichiometry, 35.743 grams of Fe₂O₃ is formed from 25.0 g of iron and an excess of oxygen.
Reaction stoichiometry
In first place, the balanced reaction is:
4 Fe + 3 O₂ → 2 Fe₂O₃
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:
- Fe: 4 moles
- O₂: 3 moles
- Fe₂O₃: 2 moles
The molar mass of the compounds is:
- Fe: 55.85 g/mole
- O₂: 32 g/mole
- Fe₂O₃: 159.7 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Fe: 4 moles ×55.85 g/mole= 223.4 grams
- O₂: 3 moles ×32 g/mole= 96 grams
- Fe₂O₃: 2 moles ×159.7 g/mole= 319.4 grams
Mass of Fe₂O₃ formed
The following rule of three can be applied: if by reaction stoichiometry 223.4 grams of Fe form 319.4 grams of Fe₂O₃, 25 grams of Fe form how much mass of Fe₂O₃?
[tex]mass of Fe_{2} O_{3}= \frac{25 grams of Fex319.4 grams of Fe_{2} O_{3}}{223.4 grams of Fe}[/tex]
mass of Fe₂O₃= 35.743 grams
Finally, 35.743 grams of Fe₂O₃ is formed from 25.0 g of iron and an excess of oxygen.
Learn more about the reaction stoichiometry:
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