How many grams of iron oxide are produced from 15.OL of oxygen at STP? (Fe203 = 159.7g/mol) %3D 2 Fe203 → 4 Fe + 3 02

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### Chemical Reaction Calculation

**Question:**
How many grams of iron oxide are produced from 15.0 L of oxygen at STP? (Fe₂O₃ = 159.7 g/mol)

**Given Reaction:**
\[2 \text{Fe}_2\text{O}_3 \rightarrow 4 \text{Fe} + 3 \text{O}_2\]

### Explanation:

1. **Understanding STP:**
   - STP stands for Standard Temperature and Pressure, which is 0°C (273.15 K) and 1 atm pressure.
   - At STP, 1 mole of any gas occupies 22.4 liters.

2. **Finding Moles of O₂:**
   - Volume of oxygen given = 15.0 L
   - Using the molar volume at STP:
   \[
   \text{Moles of O}_2 = \frac{\text{Volume}}{\text{Molar Volume}} = \frac{15.0 \, \text{L}}{22.4 \, \text{L/mol}} = 0.669 \, \text{mol}
   \]

3. **Using the Stoichiometric Coefficients:**
   - According to the reaction:
     - 3 moles of O₂ produce 2 moles of Fe₂O₃.
   - Therefore, the moles of Fe₂O₃ produced can be calculated by:
   \[
   \text{Moles of Fe}_2\text{O}_3 = \frac{2}{3} \times \text{Moles of O}_2 = \frac{2}{3} \times 0.669 \, \text{mol} = 0.446 \, \text{mol}
   \]

4. **Calculating the Mass of Fe₂O₃:**
   - Molecular weight of Fe₂O₃ = 159.7 g/mol
   
   Mass of Fe₂O₃ can be found as:
   \[
   \text{Mass of Fe}_2\text{O}_3 = \text{Moles} \times \text{Molar Mass} = 0.446 \, \text{mol} \times 159.7 \, \text{g/mol} = 71.25 \, \text{g}
   \]

###
Transcribed Image Text:### Chemical Reaction Calculation **Question:** How many grams of iron oxide are produced from 15.0 L of oxygen at STP? (Fe₂O₃ = 159.7 g/mol) **Given Reaction:** \[2 \text{Fe}_2\text{O}_3 \rightarrow 4 \text{Fe} + 3 \text{O}_2\] ### Explanation: 1. **Understanding STP:** - STP stands for Standard Temperature and Pressure, which is 0°C (273.15 K) and 1 atm pressure. - At STP, 1 mole of any gas occupies 22.4 liters. 2. **Finding Moles of O₂:** - Volume of oxygen given = 15.0 L - Using the molar volume at STP: \[ \text{Moles of O}_2 = \frac{\text{Volume}}{\text{Molar Volume}} = \frac{15.0 \, \text{L}}{22.4 \, \text{L/mol}} = 0.669 \, \text{mol} \] 3. **Using the Stoichiometric Coefficients:** - According to the reaction: - 3 moles of O₂ produce 2 moles of Fe₂O₃. - Therefore, the moles of Fe₂O₃ produced can be calculated by: \[ \text{Moles of Fe}_2\text{O}_3 = \frac{2}{3} \times \text{Moles of O}_2 = \frac{2}{3} \times 0.669 \, \text{mol} = 0.446 \, \text{mol} \] 4. **Calculating the Mass of Fe₂O₃:** - Molecular weight of Fe₂O₃ = 159.7 g/mol Mass of Fe₂O₃ can be found as: \[ \text{Mass of Fe}_2\text{O}_3 = \text{Moles} \times \text{Molar Mass} = 0.446 \, \text{mol} \times 159.7 \, \text{g/mol} = 71.25 \, \text{g} \] ###
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