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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.
<h3>Reaction stoichiometry</h3>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
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?
<u><em>mass of Fe= 11.54 grams</em></u>
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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