To determine how many grams of iron (III) oxide are formed when 25 grams of iron reacts completely with oxygen, we need to use the balanced chemical equation for the reaction and perform stoichiometry calculations.
The balanced chemical equation for the formation of iron (III) oxide is:
4 Fe + 3 O₂ → 2 Fe₂O₃
First, find the molar mass of the substances involved:
Iron (Fe): 55.85 g/mol
Oxygen (O₂): 2 × 16.00 = 32.00 g/mol
Iron (III) Oxide (Fe₂O₃): (2 × 55.85) + (3 × 16.00) = 159.70 g/mol
Next, convert the given mass of iron to moles:
25 g Fe × (1 mol Fe / 55.85 g Fe) = 0.4477 mol Fe
Now, use the stoichiometry from the balanced equation to find the moles of iron (III) oxide formed:
0.4477 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe) = 0.2239 mol Fe₂O₃
Finally, convert the moles of iron (III) oxide back to grams:
0.2239 mol Fe₂O₃ × (159.70 g Fe₂O₃ / 1 mol Fe₂O₃) = 35.75 g Fe₂O₃
Therefore, 35.75 grams of iron (III) oxide are formed when 25 grams of iron reacts completely with oxygen.
2. How many grams of iron (III) oxide are formed when 25 grams of iron reacts completely with oxygen?
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