Answer:
6.65 grams of Fe₂O₃ are required to produce 4.65g Fe.
Step-by-step explanation:
The balanced reaction is:
Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂
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:
- Fe₂O₃: 1 mole
- CO: 3 moles
- Fe: 2 moles
- CO₂: 3 moles
The molar mass of each compound is:
- Fe₂O₃: 159.7 g/mole
- CO: 28 g/mol
- Fe: 55.85 g/mole
- CO₂: 44 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Fe₂O₃: 1 mole* 159.7 g/mole= 159.7 grams
- CO: 3 moles* 28 g/mol= 84 grams
- Fe: 2 moles* 55.85 g/mole= 111.7 grams
- CO₂: 3 moles* 44 g/mole= 132 grams
Then you can apply the following rule of three: if by stoichiometry 111.7 grams of Fe are produced from 159.7 grams of Fe₂O₃, 4.65 grams of Fe are produced from how much mass of Fe₂O₃?
![mass of Fe_(2) O_(3) =(4.65 grams of Fe*159.7 grams of Fe_(2) O_(3))/(111.7grams of Fe)](https://img.qammunity.org/2022/formulas/chemistry/college/mtjjbqj2p4ecj9h9ifx1o7kwio3q61hnzw.png)
mass of Fe₂O₃= 6.65 grams
6.65 grams of Fe₂O₃ are required to produce 4.65g Fe.