Respuesta :
Answer:
1. 5.093 moles
2. 0.33 (rounded) moles
Explanation:
working out in image

Taking into account the reaction stoichiometry:
- 5.09 moles of Fe is required to react with 3.82 moles of O₂.
- 0.326 moles of O₂ are needed to produce 34.7 grams of Fe₂O₃.
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= 319.3 grams
Moles of Fe needed
The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ react with 4 moles of Fe, 3.82 moles of O₂ react with how many moles of Fe?
[tex]moles of Fe=\frac{3.82 moles of O_{2}x4 moles of Fe }{3 moles of O_{2}}[/tex]
moles of Fe= 5.09 moles
Finally, 5.09 moles of Fe is required to react with 3.82 moles of O₂.
Moles of O₂ needed
The following rules of three can be applied: If by reaction stoichiometry 319.3 grams of Fe₂O₃ are formed from 3 moles of O₂, 34.7 grams of Fe₂O₃ are formed from how many moles of O₂?
[tex]moles of O_{2} =\frac{34.7 grams of Fe_{2} O_{3} x3 moles of O_{2}}{319.3 grams of Fe_{2} O_{3}}[/tex]
moles of O₂= 0.326 moles
Finally, 0.326 moles of O₂ are needed to produce 34.7 grams of Fe₂O₃.
Learn more about the reaction stoichiometry:
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