To solve the problem, we will follow these steps:
### Step 1: Write the balanced chemical equation
The reaction between iron (Fe) and oxygen (O₂) to form iron(III) oxide (Fe₂O₃) can be represented as:
\[ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \]
### Step 2: Calculate the moles of oxygen (O₂)
Given that the mass of oxygen is 4.8 g, we can calculate the number of moles of O₂ using the formula:
\[
\text{Moles of O}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{4.8 \, \text{g}}{32 \, \text{g/mol}} = 0.15 \, \text{moles}
\]
### Step 3: Determine the moles of iron (Fe) required
From the balanced equation, we know that 3 moles of O₂ react with 4 moles of Fe. Therefore, we can find out how many moles of Fe are required for 0.15 moles of O₂:
\[
\text{Moles of Fe required} = 0.15 \, \text{moles O}_2 \times \frac{4 \, \text{moles Fe}}{3 \, \text{moles O}_2} = 0.20 \, \text{moles Fe}
\]
### Step 4: Identify the limiting reagent
We have 0.15 moles of Fe available, but we need 0.20 moles of Fe to react with 0.15 moles of O₂. Therefore, Fe is the limiting reagent.
### Step 5: Calculate the mass of oxygen consumed
Next, we need to determine how much oxygen reacts with the available iron. Since 4 moles of Fe react with 3 moles of O₂, we can find out how much O₂ is consumed by 0.15 moles of Fe:
\[
\text{Moles of O}_2 \text{ consumed} = 0.15 \, \text{moles Fe} \times \frac{3 \, \text{moles O}_2}{4 \, \text{moles Fe}} = 0.1125 \, \text{moles O}_2
\]
Now, converting this to grams:
\[
\text{Mass of O}_2 \text{ consumed} = 0.1125 \, \text{moles} \times 32 \, \text{g/mol} = 3.6 \, \text{g}
\]
### Step 6: Calculate the mass of oxygen left over
We initially had 4.8 g of O₂, and after the reaction, 3.6 g was consumed. Thus, the mass of O₂ left is:
\[
\text{Mass of O}_2 \text{ left} = 4.8 \, \text{g} - 3.6 \, \text{g} = 1.2 \, \text{g}
\]
### Step 7: Calculate the mass of iron(III) oxide (Fe₂O₃) produced
From the balanced equation, 4 moles of Fe produce 2 moles of Fe₂O₃. Therefore, for 0.15 moles of Fe:
\[
\text{Moles of Fe}_2\text{O}_3 \text{ produced} = 0.15 \, \text{moles Fe} \times \frac{2 \, \text{moles Fe}_2\text{O}_3}{4 \, \text{moles Fe}} = 0.075 \, \text{moles Fe}_2\text{O}_3
\]
Now, converting this to grams:
\[
\text{Mass of Fe}_2\text{O}_3 = 0.075 \, \text{moles} \times 160 \, \text{g/mol} = 12 \, \text{g}
\]
### Conclusion
Based on the calculations:
1. Iron (Fe) is the limiting reagent.
2. The mass of oxygen left over is 1.2 g.
3. The mass of iron(III) oxide (Fe₂O₃) produced is 12 g.
### Summary of Correct Statements
- Statement 1: Iron is the limiting reagent (Correct)
- Statement 2: The mass of oxygen left over is 1.2 g (Correct)
- Statement 3: The mass of iron oxide produced is 12 g (Correct)
- Statement 4: Oxygen is the limiting reagent (Incorrect)
