Answer:
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?
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?
moles of O₂= 0.326 moles
Finally, 0.326 moles of O₂ are needed to produce 34.7 grams of Fe₂O₃.
Step-by-step explanation: