Approximately 25.21 grams of iron can be produced.
To find the amount of iron (Fe) produced in the reaction, we first need to determine the limiting reactant. The balanced chemical equation for the reaction between magnesium (Mg) and iron(III) oxide (Fe2O3) is:
\[ 3\text{Mg} + \text{Fe}_2\text{O}_3 \rightarrow 2\text{Fe} + 3\text{MgO} \]
1. **Convert Mass to Moles:**
\[ \text{Moles of Mg} = \frac{17.25 \, \text{g}}{24.31 \, \text{g/mol}} \]
\[ \text{Moles of Fe}_2\text{O}_3 = \frac{17.25 \, \text{g}}{159.69 \, \text{g/mol}} \]
2. **Determine the Limiting Reactant:**
- The balanced equation's stoichiometry indicates that 3 moles of Mg react with 1 mole of Fe2O3.
- The mole ratio of Mg to Fe2O3 is \(3:1\).
- The moles of Mg and Fe2O3 are equal, so Mg is the limiting reactant.
3. **Calculate Moles of Fe Produced:**
- Using the stoichiometry from the balanced equation, 2 moles of Fe are produced for every 3 moles of Mg.
- \[ \text{Moles of Fe} = \frac{2}{3} \times \text{Moles of Mg} \]
4. **Convert Moles to Grams:**
- \[ \text{Grams of Fe} = \text{Moles of Fe} \times \text{Molar Mass of Fe} \]
Calculate and substitute the values to find the final answer. The molar mass of Fe is approximately 55.85 g/mol.
\[ \text{Grams of Fe} = \frac{2}{3} \times \left( \frac{17.25 \, \text{g}}{24.31 \, \text{g/mol}} \right) \times 55.85 \, \text{g/mol} \]
\[ \text{Grams of Fe} \approx 25.21 \, \text{g} \]
Therefore, approximately 25.21 grams of iron can be produced.