If 100g of iron reacts with excess oxygen how much rust should be produced

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:

1. **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:
- \(2 \times 55.85\, \text{g/mol} + 3 \times 16.00\, \text{g/mol} = 159.7\, \text{g/mol}\)

2. **Moles of Iron:**
- Calculate the number of moles of iron in 100 g:
\[
\text{Moles of Fe} = \frac{100\, \text{g}}{55.85\, \text{g/mol}} \approx 1.79\, \text{moles of Fe}
\]

3. **Using Stoichiometry:**
- According to the balanced equation, 4 moles of Fe produce 2 moles of Fe₂O₃:
\[
1 \text{ mole of Fe produces } \frac{2}{4} \text{ or } \frac{1}{2} \text{ moles of Fe}_2\text{O}_3
\]
\[
1.79 \text{ moles of Fe produces } 1.79 \times \frac{1}{2} = 0.895 \text{ moles of Fe}_2\text{O}_3
\]

4. **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.