To solve the problem, we need to determine the minimum amount of \( \text{Mn}_3\text{O}_4 \) and aluminum (Al) required to produce at least 612 g of \( \text{Al}_2\text{O}_3 \) and 825 g of manganese (Mn).
### Step-by-Step Solution:
1. **Write the Balanced Chemical Equation:**
The reaction between manganese(IV) oxide and aluminum can be represented as:
\[
\text{Mn}_3\text{O}_4 + 4\text{Al} \rightarrow 3\text{Mn} + 2\text{Al}_2\text{O}_3
\]
2. **Calculate Moles of \( \text{Al}_2\text{O}_3 \):**
The molar mass of \( \text{Al}_2\text{O}_3 \) is approximately 102 g/mol. To find the moles of \( \text{Al}_2\text{O}_3 \):
\[
\text{Moles of } \text{Al}_2\text{O}_3 = \frac{612 \, \text{g}}{102 \, \text{g/mol}} = 6 \, \text{moles}
\]
3. **Calculate Moles of Mn:**
The molar mass of Mn is approximately 55 g/mol. To find the moles of Mn:
\[
\text{Moles of Mn} = \frac{825 \, \text{g}}{55 \, \text{g/mol}} = 15 \, \text{moles}
\]
4. **Determine Moles of \( \text{Mn}_3\text{O}_4 \) Required for \( \text{Al}_2\text{O}_3 \):**
From the balanced equation, for every 2 moles of \( \text{Al}_2\text{O}_3 \), 1 mole of \( \text{Mn}_3\text{O}_4 \) is required. Therefore, for 6 moles of \( \text{Al}_2\text{O}_3 \):
\[
\text{Moles of } \text{Mn}_3\text{O}_4 = \frac{6 \times 1}{2} = 3 \, \text{moles}
\]
5. **Determine Moles of Aluminum Required for \( \text{Al}_2\text{O}_3 \):**
From the balanced equation, for every 2 moles of \( \text{Al}_2\text{O}_3 \), 4 moles of Al are required. Therefore, for 6 moles of \( \text{Al}_2\text{O}_3 \):
\[
\text{Moles of Al} = \frac{6 \times 4}{2} = 12 \, \text{moles}
\]
6. **Determine Moles of \( \text{Mn}_3\text{O}_4 \) Required for Mn:**
From the balanced equation, for every 3 moles of Mn, 1 mole of \( \text{Mn}_3\text{O}_4 \) is required. Therefore, for 15 moles of Mn:
\[
\text{Moles of } \text{Mn}_3\text{O}_4 = \frac{15 \times 1}{3} = 5 \, \text{moles}
\]
7. **Determine Moles of Aluminum Required for Mn:**
From the balanced equation, for every 3 moles of Mn, 4 moles of Al are required. Therefore, for 15 moles of Mn:
\[
\text{Moles of Al} = \frac{15 \times 4}{3} = 20 \, \text{moles}
\]
8. **Determine the Minimum Moles Required:**
- For \( \text{Mn}_3\text{O}_4 \): The maximum required is 5 moles (from Mn).
- For Al: The maximum required is 20 moles (from Mn).
9. **Calculate Mass of \( \text{Mn}_3\text{O}_4 \):**
The molar mass of \( \text{Mn}_3\text{O}_4 \) is approximately 229 g/mol. Thus, the mass required is:
\[
\text{Mass of } \text{Mn}_3\text{O}_4 = 5 \, \text{moles} \times 229 \, \text{g/mol} = 1145 \, \text{g}
\]
10. **Calculate Mass of Aluminum:**
The molar mass of Al is approximately 27 g/mol. Thus, the mass required is:
\[
\text{Mass of Al} = 20 \, \text{moles} \times 27 \, \text{g/mol} = 540 \, \text{g}
\]
### Final Answer:
The minimum amount of \( \text{Mn}_3\text{O}_4 \) and Al required is:
- \( \text{Mn}_3\text{O}_4 \): 1145 g
- Al: 540 g
