4Fe(s) + 3O₂(g) ⟶ 2Fe₂O₃(s)
How many grams of iron (III) oxide are formed when 25 grams of iron reacts completely with oxygen?
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.