First, we need to determine the stoichiometric ratio of iron (Fe) to iron (III) oxide (Fe₂O₃) in the balanced chemical equation:
4 moles of Fe react to form 2 moles of Fe₂O₃.
Now, we need to convert the given mass of iron (25 g) to moles, using its molar mass (55.85 g/mol):
(25 g) / (55.85 g/mol) = 0.447 moles of Fe.
Since 4 moles of Fe react to form 2 moles of Fe₂O₃, we can find the moles of Fe₂O₃ formed using the stoichiometric ratio:
(0.447 moles of Fe) x (2 moles of Fe₂O₃ / 4 moles of Fe) = 0.2235 moles of Fe₂O₃.
Now, we convert the moles of Fe₂O₃ into grams using its molar mass (159.69 g/mol):
(0.2235 moles of Fe₂O₃) x (159.69 g/mol) = 35.69 g of Fe₂O₃.
So, 35.69 grams of iron (III) oxide are formed when 25 grams of iron reacts completely with oxygen.
4Fe(s) + 3O₂(g) ⟶ 2Fe₂O₃(s)
How many grams of iron (III) oxide are formed when 25 grams of iron reacts completely with oxygen?
1 answer