Taking into account the reaction stoichiometry, 11.54 grams of Fe are formed from 16.5 grams of iron(III) oxide if hydrogen gas is in excess.
Reaction stoichiometry
In first place, the balanced reaction is:
Fe₂O₃ + 3 H₂ → 2 Fe + 3 H₂O
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
- H₂: 3 moles
- Fe: 2 moles
- H₂O: 3 moles
The molar mass of the compounds is:
- Fe₂O₃: 159.7 g/mole
- H₂: 2 g/mole
- Fe: 55.85 g/mole
- H₂O: 18 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
- H₂: 3 moles ×2 g/mole= 6 grams
- Fe: 2 moles ×55.85 g/mole= 111.7 grams
- H₂O: 3 moles ×18 g/mole= 54 grams
Mass of each product formed
The following rule of three can be applied: if by reaction stoichiometry 159.7 grams of Fe₂O₃ form 111.7 grams of Fe, 16.5 grams of Fe₂O₃ form how much mass of Fe?
[tex]mass of Fe=\frac{16.5 grams of Fe_{2} O_{3} x111.7 grams of Fe}{159.7 grams of Fe_{2} O_{3}}[/tex]
mass of Fe= 11.54 grams
Then, 11.54 grams of Fe are formed from 16.5 grams of iron(III) oxide if hydrogen gas is in excess.
Learn more about the reaction stoichiometry:
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