Rust is primarily composed of iron(III) oxide (Fe₂O₃•nH₂O), but for simplicity, let's consider the formation of iron(III) oxide (Fe₂O₃) in the reaction with oxygen. The balanced chemical reaction for the formation of rust (Fe₂O₃) can be represented as:
\[ 4Fe + 3O_2 \rightarrow 2Fe_2O_3 \]
From the stoichiometry of this reaction:
- 4 moles of iron (Fe) react with 3 moles of oxygen (O₂) to produce 2 moles of iron(III) oxide (Fe₂O₃).
Now, let's calculate how much rust (Fe₂O₃) is produced from 100g of iron:
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Molar Mass Calculation:
- The molar mass of iron (Fe) is approximately 55.85 g/mol.
- The molar mass of iron(III) oxide (Fe₂O₃) is calculated as follows:
- \(2 \times 55.85, \text{g/mol} + 3 \times 16.00, \text{g/mol} = 159.7, \text{g/mol}\)
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Moles of Iron:
- Calculate the number of moles of iron in 100 g: \[ \text{Moles of Fe} = \frac{100, \text{g}}{55.85, \text{g/mol}} \approx 1.79, \text{moles of Fe} \]
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Using Stoichiometry:
- According to the balanced equation, 4 moles of Fe produce 2 moles of Fe₂O₃: \[ 1 \text{ mole of Fe produces } \frac{2}{4} \text{ or } \frac{1}{2} \text{ moles of Fe}_2\text{O}_3 \] \[ 1.79 \text{ moles of Fe produces } 1.79 \times \frac{1}{2} = 0.895 \text{ moles of Fe}_2\text{O}_3 \]
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Mass of Rust Produced:
- Calculate the mass of Fe₂O₃ produced: \[ \text{Mass of Fe}_2\text{O}_3 = 0.895, \text{moles} \times 159.7, \text{g/mol} \approx 142.7, \text{g} \]
Therefore, approximately 142.7 grams of rust (Fe₂O₃) would be produced from 100 grams of iron reacting with excess oxygen.
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