To solve the problem, we need to determine the minimum amounts of `Mn3O4` and `Al` required to produce at least `612 g` of `Al2O3` and `825 g` of `Mn`.
### Step-by-Step Solution:
1. **Write the Balanced Chemical Reaction:**
The reaction between `Mn3O4` and `Al` can be represented as:
\[
3 \text{Mn}_3\text{O}_4 + 8 \text{Al} \rightarrow 9 \text{Mn} + 4 \text{Al}_2\text{O}_3
\]
2. **Calculate Moles of `Al2O3`:**
The molar mass of `Al2O3` is approximately `102 g/mol`. To find the minimum moles of `Al2O3` needed:
\[
\text{Moles of } Al2O3 = \frac{612 \, \text{g}}{102 \, \text{g/mol}} = 6 \, \text{moles}
\]
3. **Calculate Moles of `Mn`:**
The molar mass of `Mn` is approximately `55 g/mol`. To find the minimum moles of `Mn` needed:
\[
\text{Moles of } Mn = \frac{825 \, \text{g}}{55 \, \text{g/mol}} = 15 \, \text{moles}
\]
4. **Determine Moles of `Mn3O4` Required for `Al2O3`:**
From the balanced equation, `4 moles` of `Al2O3` are produced from `3 moles` of `Mn3O4`. Therefore, for `6 moles` of `Al2O3`:
\[
\text{Moles of } Mn3O4 = \frac{6 \, \text{moles} \times 3 \, \text{moles of } Mn3O4}{4 \, \text{moles of } Al2O3} = 4.5 \, \text{moles}
\]
5. **Determine Moles of `Mn3O4` Required for `Mn`:**
From the balanced equation, `9 moles` of `Mn` are produced from `3 moles` of `Mn3O4`. Therefore, for `15 moles` of `Mn`:
\[
\text{Moles of } Mn3O4 = \frac{15 \, \text{moles} \times 3 \, \text{moles of } Mn3O4}{9 \, \text{moles of } Mn} = 5 \, \text{moles}
\]
6. **Determine the Limiting Reactant:**
We have calculated:
- `4.5 moles` of `Mn3O4` required for `Al2O3`
- `5 moles` of `Mn3O4` required for `Mn`
The larger amount (`5 moles`) will dictate the amount of `Mn3O4` required.
7. **Calculate the Mass of `Mn3O4`:**
The molar mass of `Mn3O4` is approximately `229 g/mol`. Therefore, the mass required is:
\[
\text{Mass of } Mn3O4 = 5 \, \text{moles} \times 229 \, \text{g/mol} = 1145 \, \text{g}
\]
8. **Calculate the Moles of `Al` Required:**
From the balanced equation, `8 moles` of `Al` are required for `3 moles` of `Mn3O4`. Therefore, for `5 moles` of `Mn3O4`:
\[
\text{Moles of } Al = \frac{5 \, \text{moles} \times 8 \, \text{moles of } Al}{3 \, \text{moles of } Mn3O4} = 13.33 \, \text{moles}
\]
9. **Calculate the Mass of `Al`:**
The molar mass of `Al` is approximately `27 g/mol`. Therefore, the mass required is:
\[
\text{Mass of } Al = 13.33 \, \text{moles} \times 27 \, \text{g/mol} = 360 \, \text{g}
\]
### Final Answer:
The minimum amounts required are:
- Mass of `Mn3O4`: **1145 g**
- Mass of `Al`: **360 g**
